Class 12 - Physics - Atoms

**Question1.**

Choose the correct alternative from the clues given at the end of the each statement:

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model (much greater than/no different from/much less than.)

(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force (Thomson’s model/ Rutherford’s model.)

(c) A classical atom based on .......... is doomed to collapse (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in.......... (Thomson’s model/ Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in.......... (Rutherford’s model/both the models.)

Answer:

(a) The sizes of the atom in Thomson’s model is ** no different from** the atomic size in Rutherford’s model.

(b) In the ground state of ** Thomson’s model** electrons are in stable equilibrium, while in

(c) A classical atom based on ** Rutherford’s model **is doomed to collapse (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a ** Thomson’s model** but has a highly non-uniform mass distribution in

(e) The positively charged part of the atom possesses most of the mass in ** both**.

** Question2.**

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil.

(Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Answer:

In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough.

This is because the mass of hydrogen (1.67 × 10^{−27} kg) is less than the mass of incident α−particles (6.64 × 10^{−27} kg).

Thus, the mass of the scattering particle is more than the target nucleus (hydrogen).

As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment

**Question 3.**

What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

Using Rydberg’s formula:

(hc/ λ) = (21.76 x 10^{-19}) ((1/n_{1}^{2})-(1/n_{2}^{2}))

Where h = Planck’s constant = 6.6 × 10^{−34} Js, c = Speed of light = 3 × 10^{8} m/s,

R_{H} = (21.76 x 10^{-19}), (n_{1} and n_{2} are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values n_{1} = 3 and n_{2} = ∞.

(hc/ λ) = (21.76 x 10^{-19}) [(1)/ (3^{2}) - (1)/ (∞^{2})]

λ = (6.6 x 10^{-34} x 3 x 10^{8} x9)/ (21.76 x 10^{-19})

λ =8.189 x 10^{-7} m

=818.9nm

**Question 4.**

A difference of 2.3 eV separates two energy levels in an atom.

What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Answer:

Given:

E = 2.3 eV

= 2.3 x 1.6 x 10^{-19} J

=3.68 x 10^{-19} J

As E = (hf)

Therefore, h = (E/h)

= (3.68 x 10^{-19})/ (6.62 x 10^{-34})

=5.6 x 10^{14} Hz

**Question 5.**

The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Answer:

Given:

Ground state energy of hydrogen atom, E = − 13.6 eV

This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy is equal to the negative of two times of kinetic energy.

Potential energy = − (2 × (13.6)) = − 27 .2 eV

**Question 6.**

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level.

Determine the wavelength and frequency of photon.

Answer:

Given:

For ground level, n_{1 }= 1

Energy of n_{1} level = E_{1}

E_{1} = (-13.6)/ (n_{1}^{2}) eV

= (-13.6) (1^{2})

=-13.6 eV

The atom is excited to a higher level, n_{2} = 4.

Energy of n_{2} level = E_{2}

E_{2} = (-13.6)/ (n_{2}^{2}) eV

= (-13.6) (4^{2})

E_{2}= (-13.6)/ (16) eV

The amount of energy absorbed by the photon is given as:

E = (E_{2} − E_{1})

= (-13.6/16) – (-13.6/1)

= ((13.6 x 15)/ (16)) x (1.6 x 10^{-19})

=2.04 x 10^{-18 }J

For a photon of wavelength λ, the expression of energy is written as:

E = (hc)/ (λ)

Where,

h = Planck’s constant = 6.6 × 10^{−34} Js

c = Speed of light = 3 × 10^{8} m/s

Therefore, λ = (hc)/ (E)

= (6.6 × 10^{−34} x 3 x 10^{8})/ (2.04 x 10^{-18})

= (9.7 x 10^{-8}) m

= 97nm

And, frequency of a photon is given by the relation,

ν = (c/ λ)

= (3 x 10^{8})/ (9.7 x 10^{-8})

=3.1 x 10^{15} Hz

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10^{15} Hz.

** Question 7.**

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

(b) Calculate the orbital period in each of these levels.

Answer:

Given:

Let electron of hydrogen atom at ground state n_{1 }= 1

Orbital speed of the electron = v_{1}

Charge of an electron = e

v_{1} = (e^{2})/ (n_{1} 4∏ε_{0 }(h/2∏))

= (e^{2})/ (2ε_{0}h)

Where e = 1.6 x 10^{-19}C,

ε_{0 }= Permittivity of free space = (8.85 x 10^{-12}) N^{-1}C^{2}m^{-2}

Planck’s constant h = 6.62 x 10^{-34} Js

Therefore, ν = (1.6 x 10^{-19})^{2}/ (2x 8.85 x 10^{-12}x6.62 x 10^{-34})

=0.0218 x 10^{8}

= (2.18 x 10^{6}) m/s

For level n_{2} = 2, we can write the relation for the corresponding orbital speed as:

v_{2} = (e^{2})/ (n_{2}2 ε_{0}h)

= (1.6 x 10^{-19})^{2}/ (2 x 2x 8.85 x 10^{-12}x6.62 x 10^{-34}))

= (1.09 x 10^{6}) m/s

And, for n_{3} = 3, we can write the relation for the corresponding orbital speed as:

v_{3} = (e^{2})/ (n_{3}2 ε_{0}h)

= ((1.6 x 10^{-19})^{2}/ (3 x 2 x 8.85 x 10^{-12}x6.62 x 10^{-34}))

= (7.27 x 10^{5}) m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 ×10^{6} m/s, 1.09 × 10^{6} m/s, 7.27 × 10^{5} m/s respectively.

(b) Let T_{1} be the orbital period of the electron when it is in level n_{1} = 1.

Orbital period is related to orbital speed as:

T_{1} = (2∏r_{1})/ (v_{1})

Where, r_{1} = Radius of the orbit = ((n_{1})^{2}h^{2} ε_{0})/ (∏me^{2})

h = Planck’s constant = 6.62 × 10^{−34} Js

e = Charge on an electron = 1.6 × 10^{−19} C

ε_{0} = Permittivity of free space = 8.85 × 10^{−12} N^{−1} C^{2 }m^{−2}

m = Mass of an electron = 9.1 × 10^{−31} kg

Therefore, T_{1} = (2∏r_{1})/ (v_{1})

= ((2∏x (1)^{2} x (6.62 x 10^{-34})^{2}x (8.85 x 10^{-12}))/ ((2.18 ×10^{6}) x ∏ x (9.1 × 10^{−31}) x (1.6 × 10^{−19}))

=15.27 x 10^{-17}

T_{1 }=1.527 x 10^{-16} sec

For level n_{2} = 2, we can write the period as:

T_{2} = (2∏r_{2})/ (v_{2})

Where, r_{2} = Radius of the electron in n_{2} = 2

= ((n_{2})^{2}h^{2} ε_{0})/ (∏me^{2})

Therefore, T_{2} = (2∏r_{2})/ (v_{2})

= ((2∏x (2)^{2} x (6.62 x 10^{-34})^{2}x (8.85 x 10^{-12}))/ ((1.09 ×10^{6}) x ∏ x (9.1 × 10^{−31}) x (1.6 × 10^{−19}))

=1.22 x 10-15sec

And, for level n_{3} = 3, we can write the period as:

T_{3} = (2∏r_{3})/ (v_{3})

Where, r_{3} = Radius of the electron in n_{3} = 3

= ((n_{3})^{2}h^{2} ε_{0})/ (∏me^{2})

Therefore, T_{3} = (2∏r_{3})/ (v_{3})

= ((2∏x (3)^{2} x (6.62 x 10^{-34})^{2}x (8.85 x 10^{-12}))/ ((7.27 ×10^{6}) x ∏ x (9.1 × 10^{−31}) x (1.6 × 10^{−19}))

=4.12 x 10^{-15}sec

Hence, the orbital period in each of these levels is 1.52 × 10^{−16} s, 1.22 × 10^{−15} s, and 4.12 × 10^{−15} s respectively.

**Question 8.**

The radius of the innermost electron orbit of a hydrogen atom is 5.3×10^{–11} m. What are the radii of the n = 2 and n =3 orbits?

Answer:

Given:

The radius of the innermost orbit of a hydrogen atom, r_{1} = 5.3 × 10^{−11} m.

Let r_{2} = radius of the orbit at n = 2.

It is related to the radius of the innermost orbit as:

r_{2} = (n)^{ 2} r_{1}

= (4 x 5.3 x 10^{-11})

=2.12 x 10^{-10}m

For n = 3, we can write the corresponding electron radius as:

r_{3} = (n)^{ 2} r_{1}

= (9 x 5.3 x 10^{-11})

= 4.77 x 10^{-10} m

Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10^{−10} m and 4.77 × 10^{−10} m respectively.

**Question 9.**

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer:

Given:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature 12.5 eV.

Also, the energy of the gaseous hydrogen in its ground state at room temperature = −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes= (−13.6 + 12.5) eV = −1.1 eV.

Orbital energy is related to orbit level (n) as:

E =-(13.6)/ (n^{2}) eV

For n=3, E = - (13.6)/ (9)

=-1.5eV

This energy is approximately equal to the energy of gaseous hydrogen.

It can be seen that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

Using relation for wave number for Lyman series as:

(1/ λ) = R_{H} ((1/1^{2}) – (1/n^{2}))

Where,

R_{H} = Rydberg constant = 1.097 × 10^{7} m^{−1}

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

(1/ λ) = (1.097 × 10^{7}) ((1) – (1/9))

= (1.097 x 10^{7} x (8/9))

λ = (9)/ (8 x 1.097 x 10^{7}) = 102.55nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

(1/ λ) = (1.097 x 10^{7}) x ((1/1^{2}) – (1/2^{2}))

= (1.097 x 10^{7}) (1-(1/4))

= (1.097 x 10^{7}) (3/4)

λ = (4)/ (3x1.097 x 10^{7})

=121.54nm

For n = 3,

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

(1/ λ) = (1.097 x 10^{7}) x ((1/2^{2}) – (1/3^{2}))

= (1.097 x 10^{7}) x ((1/4) – (1/9))

= (1.097 x 10^{7}) x (5/36)

λ = (36)/ (1.097 x 10^{7} x 5)

=656.33nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted.

And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

**Question 10.**

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an

orbit of radius 1.5 × 10^{11} m with orbital speed 3 × 10^{4} m/s. (Mass of earth = 6.0 × 10^{24} kg.)

Answer:

Given:

Radius of the orbit of the Earth around the Sun, r = 1.5 × 10^{11} m

Orbital speed of the Earth, ν = 3 × 10^{4} m/s

Mass of the Earth, m = 6.0 × 10^{24} kg

Angular momentum of earth = mvr = (6.0 × 10^{24} x 3 x10^{4} x 1.5 x 10^{11})

= (2.7 x 10^{40}) kg/m

According to Bohr’s model, angular momentum of earth around sun:

mvr = n x (h/2∏)

Where,

h = Planck’s constant = 6.62 × 10^{−34} Js

n = Quantum number

Or (2.7 x 10^{40}) = n x (h/2∏)

Therefore n = ((2∏) / (h)) x (2.7 x 10^{40})

= ((2∏)/ (6.62 × 10^{−34})) x (2.7 x 10^{40})

n = 2.57 x 10^{74}

Hence, the quanta number that characterizes the Earth’ revolution is

2.57 × 10^{74}

.