Class 11 - Chemistry - Structure of Atom

Question 2.1

- Calculate the number of electrons which will together weigh one gram.
- Calculate the mass and charge of one mole of electrons.

Answer.

- Mass of electron = 9.1 x 10
^{-28}g

9.1 x 10^{-28} g is mass of 1 electron.

1 g is the mass of = (1)/(9.1 x 10^{-28} )= 1.098 x 10^{27} electrons.

- Mass of one electron = 9.1 x 10
^{-31}kg

Mass of 1 mole of electron = (9.1 x 10^{-31}) x (6.022 x 10^{23}) = 5.48 x 10^{-7} Kg

Charge of one electron = 1.602 x 10^{-19} C

Charge of 1 mole of electrons = (1.602 x 10^{-19}) x (6.022 x 10^{23})

= 9.65 x 10^{4} C

Question 2.2

- Calculate the total number of electrons present in one mole of methane.
- Find (a) the total number and (b) the total mass of neutrons in 7 mg of
^{14} - (Assume that mass of a neutron = 1.675 x 10
^{-27}kg). - Find (a) the total number and (b) the total mass of protons in 34 mg of NH
_{3}at STP. - Will the answer change if the temperature and pressure are changed?

Answer.

- One moles of methane has 6.022 x 10
^{23}.

Electrons present in 1 molecule of methane (CH_{4}) = 6 + 4 = 10

Electrons in mole i.e. 6.022 x 10^{23} molecules of methane = (6.022 x 10^{23} )^{ }x (10) = 6.022 x 10^{24} electrons.

- Total number of carbon atoms.

Gram atomic mass of C-14 = 14 g = 14 x 10^{3} mg

14 x 10^{3} mg of C-14 have atoms = 6.022 x 10^{23}

7 mg of C-14 have atoms = (6.022 x 10^{23} ) x (7)/(14 x 10^{3}) = 3.011 x 10^{20} atoms.

Number of neutrons in one atom C-14 of carbon = 14 - 6 = 8

Number of neutrons present in 3.011 x 10^{20} atoms C-14 of carbon = 3.011 x 10^{20} x 8

= 2.408 x 10^{21} neutrons

Mass of one neutron = 1.675 x 10^{-27} kg

Mass of 2.408 x 10^{21} neutrons = (1.675 x 10^{-27}) x (2.408 x 10^{21})= (4.033 x 10^{-6} )Kg.

- Gram molecular mass of ammonia (NH
_{3}) = 17 g = 17 x 10^{3}mg

17 x 10^{3} mg of NH_{3} have molecules = 6.022 x 10^{23}

Number of protons present in one molecules of ammonia = (7 + 3) = 10

Number of protons present in 12.044 x 10^{20} molecules of ammonia = (12.044 x 10^{20} x 10) = 1.2044 x 10^{22} protons

Mass of one proton = 1.67 x 10^{-27} kg

Mass of 1.2044 x 10^{22} protons = (1.67 x 10^{-27}) x (1.2044 x 10^{22} )= 2.01 x 10^{-5 }Kg

No, it will not change upon changing temperature and pressure because only the number of protons and mass of protons are involved.

Question 2.3

How many neutrons and protons are there in the following nuclei?

^{13}C_{6}, ^{16}O_{8}, ^{24}Mg_{12}, ^{56}Fe_{26}, ^{88}Sr_{38}

Answer.

^{13}C_{6}

Atomic Number = 6

Mass Number = 13

Number of protons = 6

Number of neutrons = 13 – 6 = 7

^{16}O_{8}

Atomic Number = 8

Mass Number = 16

Number of protons = 8

Number of neutrons = 16 – 8 = 8

^{24}Mg_{12}

Atomic Number = 12

Mass Number = 24

Number of protons = 12

Number of neutrons = 24 – 12 = 12

^{56}Fe_{26}

Atomic Number = 26

Mass Number = 56

Number of protons = 26

Number of neutrons = 56 – 26 = 30

^{88}Sr_{38}

Atomic Number = 38

Mass Number = 88

Number of protons = 38

Number of neutrons = 88 – 38 = 50

Question 2.4

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A).

- Z = 17, A = 35
- Z = 92, A = 233
- Z = 4, A= 9

Answer.

^{ 35}C_{17}

_{233U92}

^{9}Be_{4}

Question 2.5

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm.

Calculate the frequency (v) and wave number (v) of the yellow light.

Answer.

We know that v = (c/ λ)

c = 3 x 10^{8} m/s

λ = 580 nm = 580 x 10^{-9} m

v = (3 x 10^{8}) / (580 x 10^{-9}) = 5.17 x 10^{14} s^{-1}

Wave number = 1/ λ = 1/ (580 x 10^{-9}) = 1.724 x 10^{6} m^{-1}.

Question 2.6

Find energy of each of the photons which

- Corresponds to light of frequency 3 x 10
^{15} - Have wavelength of 0.50 A
^{o}.

Answer.

- Energy of photon (E) = hv

h = 6.626 x 10^{-34} J s

v = 3 x 10^{15} Hz = 3 x 10^{15} s^{-1}

E = hv = 1.986 x 10^{18} J

- Energy of photon (E) = hv =(hc/v)

h = 6.626 x 10^{34} J s

c = 3 x 10^{8} m/s

λ = 0.50 A^{o} = 0.5 x 10^{-10} m

E = (6.626 x 10^{-34} J s) x (3 x 10^{8} m/s) / (0.5 x 10^{-10 }m) = 3.98 x 10^{-15} J

Question 2.7

Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2 x 10^{-10} s.

Answer.

Frequency (v) = 1/Period = (1)/ (2 x 10^{-10} s) = 5 x 10^{9} s^{-1}

λ = (c/v) = (3 x 10^{8}) / (5 x 10^{9}) = 6 x 10^{-2} m

Wave number = (1/ λ) = (1)/(6 x 10^{-2}) = 16.66 m^{-1}

Question 2.8

What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Answer.

E = (hc)/(λ)

H = 6.626 x 10^{-34} Js

C = 3 x 10^{8} m/s

λ = 4000 pm = 4000 x 10^{-12} = 4 x 10^{-9 }m

Energy of photon, E = 4.969 x 10^{-17 }J.

4.965 x 10^{-17} J is the energy of photon = 1

1 J is the energy of photon = 1/(4.969 x 10^{-17}) = 2.012 x 10^{16} photons

Question 2.9

A photon of wavelength 4 x 10^{-7} m strikes on metal strikes on metal surface, the work function of the metal being 2.13 eV.

Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV = 1.6020 x 10^{-19} J).

Answer.

- Energy of photon (E) = hv = (hc)/(λ)

E = (6.626 x 10^{-34} Js x 3 x 10^{8} m/s)/ (4 x 10^{-7 }m) = 4.97 x 10^{-19} J

= (1 eV) x (4.97 x 10^{-19})/(1.602 x 10^{-19}) = 3.1 eV

- Kinetic energy of emission = (E – work function) =(3.1 – 2.13) = 0.97 eV
- E. of emission = ½ mv
^{2}= 0.97 eV

= 0.97 x 1.602 x 10^{-19} J = 0.97 x 1.602 x 10^{-19} kg m^{2} s^{-2}

v^{2} = (2 x 0.97 x 1.602 x 10^{-19})/ (9.1 x 10^{-31})

On solving,

v = 0.583 x 10^{6} m/s

Question 2.10

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom.

Calculate the ionization energy of sodium in kJ mol^{-1}.

Answer.

E = (hc)/(λ)

λ = 242 nm = 242 x 10^{-9} m

c = 3 x 10^{8} m/s

h = 6.626 x 10^{-34} Js

E = (6.626 x 10^{-34} Js) x (3 x 10^{8} m/s) / (242 x 10^{-9 }m) = 0.0821 x 10^{-17} J

Ionisation energy per mole (E) = [(0.0821 x 10^{-17} J) x (6.022 x 10^{23} mol^{-1} J)]/(1000) = 494 kJ/mol

Question 2.11

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second**.**

Answer.

E = hv = (hc)/(λ)

h = 6.626 x 10^{-34} J s

c = 3 x 10^{8} m/s

λ = 0.57 x 10^{-6} m

E = [(6.626 x 10^{-34} Js) x (3 x 10^{8} m/s)]/(0.57 x 10^{-6} m) = 3.48 x 10^{-19} J

Rate of emission of quanta per second = Power / Energy

P = 25 watt = 25 J/s

E = 3.48 x 10^{-19} J

E = (25)/(3.48 x 10^{-19}) = 7.18 x 10^{19 }s^{-1}

Question 2.12

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A^{o}.

Calculate threshold frequency (v_{o}) and work function (w_{o}) of the metal.

Answer.

Threshold frequency (v_{o}) = c/ λ

v_{o} = (3 x 10^{8} m/s)/(68 x 10^{-8} m) = 4.41 x 10^{14} s^{-1}

Work function, w_{o} = hv_{o} = (6.626 x 10^{-34} Js) x (4.41 x 10^{14} s^{-1})

w_{o} = 2.92 x 10^{-19} J

Question 2.13

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer.

According to Balmer formula,

Wave number = R_{H} [1/n^{2}_{1} – 1/n^{2}_{2}] cm^{-1}

n_{1} = 2

n_{2} = 4

R_{H} = 109678 cm^{-1}

λ = (1)/(wave number) = (16)/(109678 x 3) cm = 486 nm

Question 2.14

How much energy is required to ionize a H atom if the electron occupies n = 5 orbit?

Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

Answer.

Energy for a hydrogen electron present in a particular energy shell,

E_{n} = - (13.12/n^{2}) x 10^{5} J mol^{-1}

E_{n} = - (13.12 x 10^{5})/(n^{2} x 6.022 x 10^{23}) J atom^{-1}

E_{n} = - (-2.18 x 10^{-18})/n^{2} J atom^{-1}

Ionisation energy for hydrogen electron present in orbit n = 5

IE_{5} = (E _{∞} - E_{5})= 0 – [(-2.18 x 10^{-18})/25] J atom^{-1} = 8.72 x 10^{-20} J atom^{-1}

Ionisation energy for hydrogen electron present in orbit, n = 1

IE_{1} = (E_{∞} - E_{1})= 0 – [(-2.18 x 10^{-18})/1] = 2.18 x 10^{-18 }J atom^{-1}

Comparing IE_{1} and IE_{5},

(IE_{1}/IE_{5} )= 25

Energy required to remove an electron from first orbit in a hydrogen atom is 25 times the energy needed to remove an electron from fifth orbit.

Question 2.15

What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Answer.

Number of spectral lines produced when an electron in the nth level drops down to the ground state is given by n(n – 1)/2.

Given, n = 6

Number of spectral lines = (6) x (5 /2) = 15

Given by Σ (n_{2} – n_{1}) = Σ(6 – 1) = Σ5 = 5 + 4 + 3 + 2 + 1 = 15

Question 2.16

- The energy associated with the first orbit in the hydrogen atom is -2.18 x 10
^{-18}J atom^{-1}. What is the energy associated with the fifth orbit? - Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Answer.

- E
_{n}= -21.8 x 10^{-19}/n^{2}J

E_{5} = -21.8 x 10^{-18}/5^{2} = 8.72 x 10^{-20} J

- For H atom, r
_{n}= 0.529 x n^{2}A^{o}

r_{5} = 0.529 x 5^{2} = 13.225 A^{o } = 1.3225 nm

Question 2.17

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer.

For the Balmer series, n_{1} = 2

Wave number = (R)(1/2^{2} – 1/n_{2}^{2})

Wave number = (1/ λ)

For λ to be maximum, wave number should be minimum.

This can happen when n_{2} is minimum, that is, n_{2} = 3.

Wave number = (1.097 x 10^{7} m^{-1}) ((1/2^{2} – 1/3^{2}) = (1.097 x 10^{7})x (5/36) m^{-1 }

Wave number = 1.526 x 10^{6} m^{-1}

Question 2.18

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and

what is the wavelength of the light emitted when the electron returns to the ground state?

The ground state electron energy is -2.18 x 10^{11} ergs.

Answer.

1 erg = 10^{-7} J

As ground state electronic energy = - 2.18 x 10^{-11} erg

E_{n} = - 21.8 x 10^{11}/n^{2} ergs.

ΔE = E_{5} – E_{1} = 2.18 x 10^{-11} (1/1^{2} – 1/5^{2}) = 2.18 x 10^{-11} (24/25) = 2.09 x 10^{-11} erg = 2.09 x 10^{-18} J.

When electron returns to ground state (n = 1) energy emitted = 2.09 x 10^{-11} ergs.

E = hv = (hc)/(λ)

λ = (hc/E)

Put all the values in the equations,

λ = 951 x 10^{-6 }cm = 951 A^{o}

Question 2.19

The electron energy in hydrogen atom is given by E_{n} = (-2.18 x 10^{-18})/n^{2} J.

Calculate the energy required to remove an electron completely from the n = 2 orbit.

What is the longest wavelength of light in cm that can be used to cause this transition?

Answer.

ΔE = E∞ - E_{2} = 0 – (-2.18 x 10^{-18} J atom^{-1}/2^{2}) = 5.45 x 10^{-19} J atom^{-1}

ΔE = hv = (hc)/(λ)

λ = (hc)/(ΔE) = (6.626 x 10^{-34} Js) x (3 x 10^{8} m/s) / (5.45 x 10^{-19} J) = 3.647 x 10^{-7} m

= 3647 A^{0}

Question 2.20

Calculate the wavelength of an electron moving with a velocity of 2.05 x 10^{7 }m/s.

Answer.

According to de Broglie equation, λ = h/mv

λ = 6.626 x 10^{-34 }Js / (9.11 x 10^{-31} kg) (2.05 x 10^{7} m/s) = 3.55 x 10^{-11} m

Question 2.21

The mass of an electron is 9.1 x 10^{-31} kg. If its K.E. is 3.0 x 10^{-25} J, calculate its wavelength.

Answer.

K.E. = ½ mv^{2}

v = √ (2K.E./m)

= √(2 x 3 x 10^{-25}J)/(9.11 x 10^{-31} kg) = 812 m/s

By de Broglie equation, λ = (h/mv) = 6.626 x 10^{-34} Js/(9.11 x 10^{-31} kg) (812 m/s)

= 8.9625 x 10^{-7 }m

Question 2.22

Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na^{+}, K^{+}, Mg^{2+}, Ca^{2+}, S^{2-}, Ar.

Answer.

Na^{+ }and Mg^{2+} are iso-electronic species, they have 10 electrons.

K^{+}, Ca^{2+}, S^{2-} are iso-electronic species, having 18 electrons.

Question 2.23

- Write the electronic configuration of the following ions: (a) H
^{-}(b) Na^{+}(c) O^{2-}(d) F^{-}. - What are the atomic number of elements whose outermost electrons are represented by (a) 3s
^{1}(b) 2p^{3}and (c) 3p^{5}? - Which atoms are indicated by the following configurations?

(a) [He] 2s^{1} (b) [Ne] 3s^{2} 3p^{3} (c) [Ar] 4s^{2} 3d^{1}.

Answer.

- (a) 1s
^{2}(b) 1s^{2}2s^{2}2p^{6}(c) 1s^{2}2s^{2}2p^{6}(d) 1s^{2}2s^{2}2p^{6} - (a) Na (Z = 11) has outermost electronic configuration = 3s
^{1}.

(b) N (Z = 7) has outermost electronic configuration = 2p^{3}.

(c) Fe (Z = 9) has outermost electronic configuration = 3p^{5}.

(iii) (a) Li

(b) P

(c) Sc

Question 2.24

What s the lowest value of n that allows g orbitals to exist?

Answer.

The lowest value of l where ‘g’ orbital can be present is 4.

The lowest value of n where ‘g’ orbital can be present is 4 + 1 = 5.

Question 2.25

An electron is in one of the 3d orbitals. Give the possible values of n, l and m_{1} for this electron.

Answer.

For electron in 3d orbital n = 3, l = 2, m_{i} = -2, -1, 0, 1, 2.

Question 2.26

An atom of an element contains 29 electrons and 35 neutrons.

Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer.

- For a neutral atom, number of protons = number of electrons = 29
- Electronic configuration = 1s
^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1}.

Question 2.27

Give the number of electrons in the species H^{+}_{2}, H_{2} and O^{+}_{2}.

Answer.

H^{2+} = 2 – 1 = 1 electron

H_{2} = _{1}H + _{1}H = 2 electrons

O_{2}^{+}= 16 – 1 = 15 electrons

Question 2.28

- An atomic orbital has n = 3. What are the possible values of l and m
_{1}? - List the quantum numbers (m
_{1}and l) of the electrons of 3d orbital. - Which of the following orbitals are possible?

1p, 2s, 2p and 3f

Answer.

- n = 3, l = 0, 1, 2

For l = 0, m_{l} = 0

For l = 1, m_{l} = +1, 0, -1

For l = 2, m_{l} = +2, +1, 0, -1, -2

- For an electron in 3d orbital

n = 3, l = 2, m_{l} = -2, -1, 0, +1, +2

- 1p and 3f orbitals are not possible.

Question 2.29

Using s, p, d notations, describe the orbital with the following quantum numbers.

- n = 1, l = 0
- n = 3, l = 1
- n = 4, l =2
- n = 4, l = 3

Answer.

- 1s orbital
- 3p orbital
- 4d orbital
- 4f orbital

Question 2.30

Explain, giving reasons, which of the following sets of quantum numbers are not possible.

- n = 0, l = 0, m
_{1}= 0, m_{s}= +1/2 - n = 1, l = 0, m
_{1}= 0, m_{s}= -1/2 - n = 1, l = 1, m
_{1}= 0, m_{s}= +1/2 - n = 2, l = 1, m
_{1}= 0, m_{s}= -1/2 - n = 3, l = 3, m
_{1}= -3, m_{s}= +1/2 - n = 3, l = 1, m
_{1}= 0, m_{s}= +1/2

Answer.

- The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
- The set of quantum numbers is possible.
- The set of quantum numbers is not possible because for n= 1, l cannot be equal to 1. It can have zero value.
- The set of quantum numbers is not possible because for l = 0. m
_{l}cannot be +1. It must be zero. - The set of quantum numbers is not possible because for n = 3, l ≠ 3.
- The set of quantum numbers is possible.

Question 2.31

How many electrons in an atom may have the following quantum number?

- n = 4, m
_{s}= -1/2 - n = 3, l =0

Answer.

- For n = 4

Total number of electrons = 2n^{2} = 2 x 16 = 32

Half of 32 electrons will have spin quantum number m_{s} = -1/2 i.e. 16 electrons.

- n = 3

l = 0

This means it is 3s orbital which can have only 2 electrons.

Question 2.32

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength

associated with the electron revolving around the orbit.

Answer.

According to Bohr’s postulates of angular momentum,

mvr = nh/2π

2πr = nh/mv ----(i)

According to de Broglie equation,

λ = h/mv ----(ii)

Substitute the values of equation (ii) in equation (i),

2πr = n λ

Circumference of the Bohr orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated

with the electron revolving around the orbit.

Question 2.33

What transition in the hydrogen spectrum would have the same wavelength as the

Balmer transition n = 4 to n =2 of He^{2+} spectrum?

Answer.

For H-like particles, wave number = 1/wavelength = (1/ λ) = R_{H} Z^{2} [1/n^{2}_{1} – 1/n^{2}_{2}]

For He^{+} spectrum, Z = 4, n_{2} = 4, n_{1} = 2

Wave number = R_{H }x (4)[1/2^{2} – ¼^{2}] = 3R_{H}/4

For hydrogen spectrum, wave number = 3R_{H}/4 and Z =1

Wave number = R_{H} x 1(1/n^{2}_{1} – 1/n^{2}_{2})

R_{H }(1/n^{2}_{1} – 1/n^{2}_{2}) = (3R_{H}/4)

(1/n^{2}_{1} – 1/n^{2}_{2}) = ¾

This corresponds to n_{1} = 1, n_{2} = 2 in Hydrogen spectrum ans would have the same wavelength as in

Balmer series from n = 4 to n = 1 of He^{2+} spectrum.

Question 2.34

Calculate the energy required for the process

He^{+} (g) à He^{2+} (g) + e^{-}

The ionization energy for the H atom in the ground state is 2.18 x 10^{-18 }J atom^{-1}.

Answer.

Ionisation energy atom, E_{n} = (2.18 x 10^{-18} x Z^{2})/n^{2} J atom^{-1}

For H - atom, E_{n} = 2.18 x 10^{-18} x (1)^{2} J atom^{-1}

For He^{2+} ion, E_{n} = 2.18 x 10^{-18} x (2)^{2} = 8.72 x 10^{-18} J atom^{-1}

Question 2.35

If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a

straight line across length of scale of length 20 cm long.

Answer.

Given length of scale = 20 cm = 20 x 10^{7} nm = 2 x 10^{8} nm

Diameter of carbon atom = 0.15 nm

Number of carbon atoms which can be placed side by side in the scale = (2 x 10^{8} nm)/(0.15 nm) = 1.33 x 10^{9}

Question 2.36

2 x 10^{8} atoms of carbon are arranged side by side.

Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Answer.

Total length = 2.4 cm

Total number of atoms along the length = 2 x 10^{8}

Diameter of each atom = (2.4 cm)/(2 x 10^{8})= 1.2 x 10^{-8} cm

Radius of the atom = (Diameter/2) = 1.2 x 10^{-8} cm^{-2} = 0.6 x 10^{-8} cm

Question 2.37

The diameter of zinc atom is 2.6 A^{o}. Calculate (a) radius of zinc atom in pm and

(b) number of atoms present in a length of 1.6 cm

if the zinc atoms are arranged side by side lengthwise.

Answer.

- Radius of zinc atom = 2.6 A
^{o}/2 = 1.3 A^{o}= 1.3 x 10^{-10}m = 130 x 10^{-12}pm - Given length = 1.6 cm = 1.6 x 10
^{-2}m

Diameter of one atom = 2.6 A^{o}

= 2.6 x 10^{-10} m

Number of atoms present along the length = 1.6 x 10^{-2}/2.6 x 10^{-10}

= 6.153 x 10^{7}

Question 2.38

A certain particle carries 2.5 x 10^{-16} of static electric charge. Calculate the number of electrons present in it.

Answer.

Charge on one electron = 1.602 x 10^{-19 }C

Number of electrons carrying 2.5 x 10^{-16} C charge = 2.5 x 10^{-16}/1.602 x 10^{-19} = 1560C

Question 2.39

In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays.

If the static electric charge on the oil drop is -1.282 x 10^{-18} C, calculate the number of electrons present on it.

Answer.

Charge on oil droplet = -1.282 x 10^{-18} C

Charge on an electron = - 1.602 x 10^{-19} C

Number of electrons = q/e = (-1.282 x 10^{-18 }C)/(-1.602 x 10^{-19} C) = 8

Question 2.40

In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc.

Have been used to be bombarded by the a-particles.

If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Answer.

According to Rutherford’s experiment, heavy atoms have nucleus having large amount of positive charge.

If a thin foil of lighter atoms like aluminium be used in the Rutherford experiment,

this will obstruct the path of the fast moving alfa-particles to comparatively less extent.

The total number of alfa-particles deflected will be quite less and the particles deflected back will be negligible.

Question 2.41

Symbols ^{79}Br_{35} and ^{79}Br can be written, whereas symbols ^{35}Br_{79} and ^{35}Br are not acceptable. Answer briefly.

Answer.

Atomic number of bromine is 35. When writing the atomic number and mass number of an atom with the symbol,

atomic number is always written in subscript and mass number is written in super script. Hence, symbol ^{79}Br_{35} is not acceptable.

Br_{35} is not acceptable because atomic number of an element is fixed.

Mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate the mass number.

Question 2.42

An element with mass number 81 contains 31.7 % more neutrons as compared to protons. Assign the atomic symbol.

Answer.

Mass number = protons + neutrons

M = p + n = 81 (given)

Let p = x

n = x + (31.7/100)x = 1.317x

(x + 1.317 x) = 81

2.317 x = 81

X = (81/2.317) = 35

Protons = 35 = atomic number

Symbol of the element is ^{81}Br_{35}.

Question 2.43

An ion with mass number 37 possesses one unit of negative charge.

If the ion contains 11.1 % more neutrons than the electrons, find the symbol of the ion.

Answer.

Let the number of electrons in the ion = x

Number of neutrons = x + (11.1 x /100) = 1.111 x

Number of electrons in the neutral atom = (x) – 1 (ion possesses one unit of negative charge)

Number of protons = x – 1

Mass number = Number of protons + Number of neutrons

(1.111 x) + (x -1) = 37

2.111 x = 38

x = 18

Number of protons = Atomic number = (x -1) = 18 – 1 = 17

Symbol of ion is ^{35}Cl_{17}^{-1}.

Question 2.44

An ion with mass number 56 contains 3 units of positive charge and 30.4 % more neutrons than electrons.

Assign the symbol to this ion.

Answer.

Number of electrons in the ion = x

Number of neutrons = x + (30.4 x /100) = 1.304 x

Number of electrons in the neutral atom = x + 3 (ion possesses three units of positive charge)

Number of protons = x + 3

Mass number = Number of protons + Number of neutrons

1.304 x + x + 3 = 56

2.304 x = 53

X = 23

Number of protons = Atomic number = x + 3 = 23 + 3 = 26

Symbol of the ion is ^{56}Fe_{26}^{3+}.

Question 2.45

Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven

(b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer.

The increasing order of frequency is as follows:-

Radiations from FM radio < Amber light < Radiation from microwave oven < X-rays < Cosmic rays.

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

Question 2.46

Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10^{24},

calculate the power of this laser.

Answer.

E = nhv = (nhc)/(λ)

E = (5.6 x 10^{24}) x (6.626 x 10^{-34} Js x 3 x 10^{8} ms^{-1})/(337.1 x 10^{-9} )m

E = 3.3 x 10^{6} J

Question 2.47

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm,

calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s

(c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Answer.

λ = 616 nm = 616 x 10^{-9} m

- Frequency, v= (c/ λ)

v = (3 x 10^{8})/(616 x 10^{-9})

v = 4.87 x 10^{14} s^{-1}

- Velocity = 3 x 10
^{8}ms^{-1}

Distance travelled in 30 s = 30 x 3 x 10^{8} = 9 x 10^{9 }m

- E = hv = hc/ λ

E = (6.626 x 10^{-34} x 3 x 10^{8}) / 616 x 10^{-9}

E = 32.27 x 10^{-20} J

- Number of quanta in 2 J of energy = 2 J /32.27 x 10
^{-20}J = 6.2 x 10^{18}

Question 2.48

In astronomical observations, signals observed from the distant stars are generally weak.

If the photon detector receives a total 3.15 x 10^{-18 }J from the radiations of 600 nm,

calculate the number of photons received by the detector.

Answer.

Energy of one photon = hv = (hc/ λ)

E = (6.626 x 10^{-34} x 3 x 10^{8})/ 600 x 10^{-9}

E = 3.313 x 10^{-19} J

Total energy received = 3.15 x 10^{-18} J

Number of photons received = (3.15 x 10^{-18} )/(3.313 x 10^{-19} )= 9.51

Question 2.49

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of

duration nearly in the nano second range.

If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10^{15} ,

calculate the energy of the source.

Answer.

Frequency = ½ x 10^{-19} s = 0.5 x 10^{9} s^{-1}

E = nhv = 2.5 x10^{15} x 6.626 x 10^{-34} x 0.5 x 10^{9} = 8.28 x 10^{-10} J

Question 2.50

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm.

Calculate the frequency of each transition and energy difference between two excited states.

Answer.

λ_{1} = 589 nm = 589 x 10^{-9} m

v_{1} = (c/ λ_{1 }= (3 x 10^{8}/589 x 10^{-9} )= 5.093 x 10^{14} s^{-1}

λ_{2} = 589.6 nm = 589 x 10^{-9} m

v_{2} = c/ λ_{2} = (3 x 10^{8}/589.6 x 10^{-9} )m = 5.088 x 10^{14} s^{-1}

Energy difference = E_{2} – E_{1} = h(v_{2} – v_{1}) = (6.626 x 10^{-34}) x (5.093 – 5.088) x 10^{14} s^{-1}

= 3.31 x 10^{-22} J

Question 2.51

The work function for caesium atom is 1.9 eV.

Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation.

If the caesium element is irradiated with a wavelength 500 nm,

calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer.

Work function (W_{o}) = hv_{o}

v_{o} = (W_{o}/h) = 1.9 x 1.602 x 10^{-19}/6.626 x 10^{-3}

v_{o} = 4.59 x 10^{14} s^{-1}

λ_{o} = (c/v_{o})= 3 x 10^{8} ms^{-1}/ 4.59 x 10^{14} s^{-1} = 6.54 x 10^{-7} m

λ_{o} = 654 nm

K.E. of ejected electron = h(v – v_{o}) = hc (1/ λ – 1/ λ_{o})

K.E. = (6.626 x 10^{-34} x 3 x 10^{8}) x (1/500 x 10^{-9} – 1/654 x 10^{-9})

K.E. = ½ mv^{2} = 9.36 x 10^{-20} J

(½ x 9.11 x 10^{-31}) v^{2} = 9.36 x 10^{-20} kg m^{2} s^{-2}

v^{2} = 2.055 x 10^{11} m^{2} s^{-2} = 20.55 x 10^{10} m^{2} s^{-2}

v = 4.53 x 10^{5} ms^{-1}

Question 2.52

Following results are observed when sodium metal is irradiated with different wavelengths.

Calculate (a) threshold wavelength and , (b) Planck’s constant

λ (nm) 500 450 400

v x 10^{-5} (cm s^{-1}) 2.55 4.35 5.35

Answer.

Let the threshold wavelength to be λ_{o} nm = λ_{o} x 10^{-9} m

Following equation holds good for photoelectric emission

K.E. = ½ mv^{2} = h (v – v_{o})

½ mv^{2} = hv - hv_{o }

hv_{o} = hv – ½ mv^{2}

hc/ λ_{o }= hc/ λ – ½ mv^{2}

On substituting the value of λ and v from the data, we get

λ_{o} (1) = 541 nm

λ_{o} (2) = 546 nm

λ_{o} (3) = 542 nm

Threshold frequency = λ_{av} = { λ_{o} (1) + _{ }λ_{o} (2) + λ_{o} (3)}/2

= {541 + 546 + 542}/3 = 543(approx 540)

Question 2.53

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by

applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Answer.

Energy of the incident radiation = (Work function + kinetic energy of photoelectron)

E = hc/ λ = (6.626 x 10^{-34} Js x 3 x 10^{8} ms^{-1})/(256.7 x 10^{-9} m) = 7.74 x 10^{-19} J

E = 4.83 eV

The potential applied gives kinetic energy of the electron

Kinetic energy of the electron = 0.35 eV

Work function = (4.83 eV – 0.35 eV) = 4.48 eV

Question 2.54

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is

ejected out with a velocity of 1.5 x 10^{7} m s^{-1},

calculate the energy with which it is bound to the nucleus.

Answer.

Energy of incident photon = (hc/ λ)

E = (6.626 x 10^{-34} Js x 3 x 10^{8} ms^{-1})/150 x 10^{-12} m) = 13.25 x 10^{-16} J

Energy of the electron ejected = ½ mv^{2} = ½ x(9.11 x 10^{-31} kg) x (1.5 x 10^{7} ms^{-1})^{2} = 1.025 x 10^{-16} J

Energy with which the electron was bound to the nucleus = 13.25 x 10^{-16} J – 1.025 x 10^{-16} J

= 12.225 x 10^{-16} J

E = (12.225 x 10^{-16} J / 1.602 x 10^{-19} )eV = 7.63 x 10^{3} eV

Question 2.55

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and

can be represented as v = 3.29 x 10^{15} (Hz) [ 1/3^{2} – 1/n^{2}]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Answer.

v = (c/λ)

v = 3 x 10^{8} m/s/1285 x 10^{-9} m = 3.29 x 10^{15} (1/3^{2} – 1/n^{2})

(1/n^{2} )= 1/9 – (3 x 10^{8} / 1285 x 10^{-9}) x (1/3.29 x 10^{15}) = 1/25

n^{2} =25

n = 5

Radiations corresponding to 1285 nm lies in the infrared region.

Question 2.56

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 2111.6 pm.

Name the series to which this transition belongs and the region of the spectrum.

Answer.

Radius of nth orbit of H-like particles = 0.529 n^{2}/Z A^{o} = 52.9 n^{2}/Z pm

r_{1} = 1.3225 = 1322.5 pm = 52.9 n_{1}^{2}

r_{2} = 211.6 pm = 52.9 n_{2}^{2}/Z

r_{1}/r_{2} = 1322.5 pm/211.6 pm

n_{1}^{2}/n_{2}^{2} = 6.25

n_{1}/n_{2} = 2.5

n_{2} = 2, n_{1} = 5

Transition is from 5^{th} orbit to 2^{nd} orbit. It belongs to Balmer series.

˜ν = 1.097 x 10^{7} m^{-1} (1/2^{2} – 1/5^{2}) = 1.097 x 10^{7} x 21/100 m^{-1}

λ = 1/˜ν = 100 / (1.097 x 21 x 10^{7}) m = 434 x 10^{-9} m = 434 nm (It lies in visible range).

Question 2.57

Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used

for the highly magnified images of biological molecules

and other type of material. If the velocity of the electron in this microscope is 1.6 x 10^{6} ms^{-1},

calculate de Broglie wavelength associated with this electron.

Answer.

λ = (h/mv)

λ = 6.626 x 10^{-34} kg m^{2} s^{-1}/(9.1 x 10^{-31} kg) (1.6 x 10^{6} ms^{-1})

λ = 4.55 x 10^{-10} m = 455 pm

Question 2.58

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules.

If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Answer.

Mass of neutron = 1.675 x 10^{-27} kg

λ = (h/mv)

v = (h/mv)

v = 6.626 x 10^{34} kg m^{2} s^{-1}/(1.675 x 10^{-12} kg) (800 x 10^{-12} m) = 4.94 x 10^{4} ms^{-1}

Question 2.59

If the velocity of the electron in Bohr’s first orbit is 2.19 x 10^{6} m/s, calculate the de Broglie wavelength associated with it.

Answer.

λ = (h/mv)

λ = 6.626 x 10 ^{-34} kg m^{2} s^{-1} / (9.1 x 10^{-31} kg)(2.19 x 10^{6} ms^{-1})

λ = 3.32 x 10^{-10} m = 332 pm

Question 2.60

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 x 10^{5} m/s.

If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Answer.

We know that

λ = (h/mv)

λ = 6.626 x 10 ^{-34} kg m^{2} s^{-1} / (0.1 kg) (4.37 x 10^{5} ms^{-1})

λ = 1.516 x 10^{-28} m

Question 2.61

If the position of the electron is measured within an accuracy of __+ __0.002 nm,

calculate the uncertainty in the momentum of the electron.

Suppose the momentum of the electron is h/4 π _{m} x 0.05 nm, is there any problem in defining this value.

Answer.

Δx = 0.002 nm = 0.002 x 10^{-9} m = 2 x 10^{-12} m

Δx. Δp = h/4π

Δp = h/4 π Δx

Δp = (6.626 x 10^{-34} kg m^{2} s^{-1})/(4 x 3.14 (2 x 10^{-12} m))

Δp = 2.638 x 10^{-23} kg m s^{-1}

Actual momentum (p) = h/4 π x 0.05 nm

p = 1.055 x 10^{-24} kg m s^{-1}

It cannot be defined because actual magnitude of the momentum is smaller than the uncertainty.

Question 2.62

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.

If any of these combination(s) has/have the same energy lists:

- n = 4, l = 2, m
_{l}= -2, m_{s}= -1/2 - n = 3, l = 2, m
_{l}= 1, m_{s}= +1/2 - n = 4, l = 1, m
_{l}= 0, m_{s}= +1/2 - n = 3, l = 2, m
_{l}= -2, m_{s}= -1/2 - n = 3, l = 1, m
_{l}= -1, m_{s}= +1/2 - n = 4, l = 1, m
_{l}= 0, m_{s}= +1/2

Answer.

- 4d
- 3d
- 4p
- 3d
- 3p
- 4p

Question 2.63

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital 6 electrons in 3p orbital and 5 electrons in 4p orbital.

Which of these electrons experiences the lowest effective nuclear charge?

Answer.

4p electrons experience lowest effective nuclear charge because of the maximum magnitude of screening effect.

It is farther from the nucleus.

Question 2.64

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

- 2s and 3s
- 4d and 4f
- 3d and 3p

Answer.

Greater the penetration of the electron present in a particular orbital towards the nucleus more will be

the magnitude of the effective nuclear charge.

- 2s electron will experience more effective nuclear charge
- 4d electron will experience more effective nuclear charge
- 3p electron will experience more effective nuclear charge

Question 2.65

The unpaired electrons in Al and Si are present in 3p orbital.

Which electrons will experience more effective nuclear charge from the nucleus?

Answer.

Configuration of the two elements are

Al (Z = 13)

[Ne]^{10} 3s^{2}3p^{1}

Si (Z = 14)

[Ne]^{10} 3s^{2}3p^{2}

Unpaired electrons in Silicon (Si) will experience more effective nuclear charge

because the atomic number of the element Si is more than that of Al.

Question 2.66

Indicate the number of unpaired electrons in

- P (b) Si (c) Cr (d) Fe and (e) Kr

Answer.

- P (Z = 15)

[Ne]^{10} 3s^{2}3p^{3}

Number of unpaired electrons = 3

- Si (Z = 14)

[Ne]^{10} 3s^{2}3p^{2}

Number of unpaired electrons = 2

- Cr (Z = 24)

[Ar]^{18} 4s^{1}3p^{5}

Number of unpaired electrons = 6

- Fe (Z = 36)

[Ar]^{18} 4s^{2}3p^{6}

Number of unpaired electrons = 4

- Kr (Z = 36)

[Ar]^{18} 4s^{2}3p^{10}4p^{6}

Number of unpaired electrons = Nil

Question 2.67

- How many subshells are associated with n = 4?
- How many electrons will be present in the subshells having m
_{s}value of -1.2 for n = 4?

Answer.

- For n = 4

Number of sub- shells

(l = 0, l = 1, l = 2, l = 3) = 4

- Total number of orbitals which can be present = n
^{2}= 4^{2}= 16

Each orbital can have an electron with m_{s} = -1/2

Total number of electrons with m = -1/2 is 16.

.