The ground state energy of hydrogen atom is - 13 . 6 eV . If an electron makes a transition from an energy level - 1 . 51 eV to - 3 . 4 eV , calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Given, The energy of the first state is E 1 = - 3 . 4 eV and its orbit is n 1 . The energy of the second state is E 2 = - 1 . 51 eV and its orbit is n 2 . We know that, the energy of the orbit is given by, E = - 13 . 6 Z 2 n 2 eV E 2 = - 1 · 51 eV = - 13 . 6 eV n 2 2 ⇒ n 2 = 3 E 1 = - 3 . 4 eV = - 13 . 6 eV n 1 2 ⇒ n 1 = 2 The transition is happening from 3 r d orbit 2 n d orbit, so it is Balmer series. Say, the wavelength of the spectral line is λ . Then, Δ E = h c λ , where ∆ E is the difference in the energy levels. ∴ ∆ E = ( - 1 . 51 + 3 . 4 ) eV = 1 . 89 × 1 . 6 × 10 - 19 J ∴ λ = h c Δ E = 6 . 6 × 10 - 34 × 3 × 10 8 1 . 89 × 1 . 6 × 10 - 19 m = 6577 A 0

The spectrum of a star in the visible and the ultraviolet region was observed and the wavelength of some of the lines that could be identified was found to be 824 A → , 970 A → , 1120 A → , 2504 A → , 5173 A → , 6100 A → . Which of these lines cannot belong to the hydrogen atom spectrum? Given Rydberg constant R H = 1 . 03 × 10 7 m - 1 and 1 R H = 970 A → . Support your answer with suitable calculations.

Wave number of emitted radiation of hydrogen atom is given by: ν ¯ = 1 λ = R 1 n 2 2 - 1 n 1 2 For lyman series we take value of n 2 =1 and n 1 can be (2,3,4...) λ = 970 A ∘ 1 n 2 2 - 1 n 1 2 = 970 A ° 3 / 4 , 970 A ° 8 / 9 , 970 A ° 15 / 16 ,.... 970 A ° 1 Similarly, when calculated for balmer series keeping value of n 2 to be 2 and n 1 to be (2,3,4...) we get λ = 698 A ° ,5173.3A ° ,4619A ° ......3880A ° Therefore, the lines that do not belong to hydrogen spectrum are: 824 A → , 1120 A → , 2504 A → , 6100 A →

For what kinetic energy of a neutron will the associated de Broglie wavelength be 1 . 40 × 10 - 10 m ? The mass of a neutron is 1 . 675 × 10 - 27 kg and h = 6 . 63 × 10 - 34 Js .

Given, De-Broglie wavelength of the neutron = 1 . 40 × 10 - 10 m . Planck’s constant h = 6 . 63 × 10 - 34 Js . Mass of neutron m n = 1 . 675 × 10 − 27 kg . Kinetic energy (K) and velocity (v) are related as K = 1 2 m n v 2 . . . . . . . . . . . . 1 De-Broglie wavelength λ and velocity v are related as: λ = h m n v . . . . . . . . . . . . 2 Using eq (2) in eq (1), we get, K = 1 2 m n h 2 λ 2 m n 2 = h 2 2 λ 2 m n = 6 . 63 × 10 - 34 2 2 × 1 . 40 × 10 - 10 2 × 1 . 66 × 10 - 27 = 6 . 75 × 10 - 21 J = 6 . 75 × 10 - 21 1 . 6 × 10 - 19 = 4 . 219 × 10 - 2 eV Therefor the kinetic energy of the neutron is 6 . 75 × 10 - 21 J or 4 . 219 × 10 - 2 eV . Hence, it is the required answer.

The combination of fundamental quantities e 2 2 ε 0 hc ≡ α is known as the fine structure constant. Show that α is a pure number, of magnitude 1 137

α = e 2 2 ε 0 c h (Given) ⇒ α = 1 . 6 × 10 - 19 2 2 × 8 . 85 × 10 - 12 × 3 × 10 8 × 6 . 63 × 10 - 34 ⇒ α = 1 137 Simply it is coupling constant or measure the strength of electromagnetic force that govern how electrically charged elementary (electron, muon) and photon interact.

What is rest mass of photon?

By nature, photons are just energy packets which are travelling continuously. Therefore photons have zero rest mass as they are never stationary. Note-According to Einstein's quantum theory light propagates in the form of packets i.e. quanta of energy, which is called photon. The rest mass of photons is being zero. It can be shown, according to the relativity theory of light. According to the relativistic theory equation, the mass of the photon is computed as: m = m 0 1 − ( c 2 / v 2 ) m 0 = m 1 − ( c 2 / v 2 ) when v = 0 ; So m 0 = 0 Where m 0 is the rest mass of the Photon .

HARD

12th West Bengal Board

IMPORTANT

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The total energy of an electron in the first excited state of the hydrogen atom is $- 3.4 eV$ . What is the kinetic energy of the electron in this state?

Important Questions on Bohr Model, X-Ray Spectra, Wave-Particle Duality

HARD

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 3

The ground state energy of hydrogen atom is

- 13.6 eV

. If an electron makes a transition from an energy level

- 1.51 eV

- 3.4 eV,

calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 5

The spectrum of a star in the visible and the ultraviolet region was observed and the wavelength of some of the lines that could be identified was found to be

824 \vec{A}, 970 \vec{A}, 1120 \vec{A}, 2504 \vec{A}, 5173 \vec{A}, 6100 \vec{A}

. Which of these lines cannot belong to the hydrogen atom spectrum? Given Rydberg constant

R_{H} = 1.03 \times 10^{7} m^{- 1}

and

\frac{1}{R_{H}} = 970 \vec{A}

. Support your answer with suitable calculations.

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 6

The energy of an electron in n -th orbit is given by

E_{n} = - \frac{13.6}{n^{2}} eV

Determine the energy required to take an electron from ground state to the second excited state.

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 8

The energy of an electron in an excited hydrogen atom is

- 3.4 eV

. Calculate the angular momentum of the electron according to Bohr's theory. Planck's constant

h = 6.626 \times 10^{- 34} Js

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 9

In a hydrogen atom, a transition takes place from

n = 3

n = 2

orbit. Calculate the wavelength of the emitted photon. Will the photon be visible? To which spectral series will this photon belong?

(R = 1.097 \times 10^{7} m^{- 1})

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 11

Calculate the de-Broglie wavelength of a proton of kinetic energy

500 eV

. The mass of proton is

1.67 \times 10^{- 27} kg (h = 6.63 \times 10^{- 34} Js and 1 eV = 1.6 \times 10^{- 19} J)

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 12

For what kinetic energy of a neutron will the associated de Broglie wavelength be

1.40 \times 10^{- 10} m

? The mass of a neutron is

1.675 \times 10^{- 27} kg

and

h = 6.63 \times 10^{- 34} Js

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 20 - Bohr Model, X-Ray Spectra, Wave-Particle Duality>EXERCISE 20>Q 13

The combination of fundamental quantities

\frac{e^{2}}{2 ε_{0} hc} \equiv α

is known as the fine structure constant. Show that

α

is a pure number, of magnitude

\frac{1}{137}

The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. What is the kinetic energy of the electron in this state?

Important Questions on Bohr Model, X-Ray Spectra, Wave-Particle Duality

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The total energy of an electron in the first excited state of the hydrogen atom is $- 3.4 eV$ . What is the kinetic energy of the electron in this state?