The first four spectral lines in the Lyman series of a H-atom are λ = 1218 A , ∘ 1028 A ∘ , 974 . 3 A ∘ and 951 . 4 A ∘ . If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Let us take μ H as the reduced masses of electron of hydrogen and μ D as the reduced masses of electron of deuterium. We know that 1 λ = R 1 n f 2 - 1 n i 2 As n i and n f are fixed for Lyman series for hydrogen and deuterium λ ∝ 1 R or λ D λ H = R H R D .......................................(1) R R = m e e 4 8 ε 0 c h 3 = μ H e 4 8 ε 0 c h 3 R D = m e e 4 8 ε 0 c h 3 = μ D e 4 8 ε 0 c h 3 ⇒ R H R D = μ H μ D ............................................(2) From equation (i) and (ii) λ H λ D = μ H μ D ....................................................(3) If M is mass of the proton Reduced mass for hydrogen, μ H = m e 1 + m e / M ≃ m e 1 - m e M Reduced mass for deuterium, μ D = 2 M × m e 2 M 1 + m e / 2 M ≃ m e 1 - m e 2 M μ H μ D = m e 1 - m e M m e 1 - m e 2 M = 1 - m e M 1 - m e 2 M - 1 ≈ 1 - m e 2 M μ H μ D = 1 - 1 2 × 1840 = 0 . 99973 ⇒ λ D = 0 . 99973 λ H Using λ H = 1218 A , 0 1028 A , 0 974 . 3 A , 0 951 . 4 A 0 We get λ D = 1217 . 7 A , 0 1027 A , 0 974 . 04 A , 0 951 . 1 A 0 Shift in wavelength λ H - λ D = 0 . 3 A 0

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12th CBSE

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Positronium is just like a $H$ -atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Important Questions on Atoms

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12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>SA>Q 12.2

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

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12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>SA>Q 12.21

Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

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12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>SA>Q 12.22

Show that the first few frequencies of light that is emitted when electrons fall to the $n^{t h}$ level from levels higher than $n$ are approximate harmonics (i.e. in the ratio $1 : 2 : 3$ ) when $n > > 1$ .

EASY

12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>SA>Q 12.23

What is the minimum energy that must be given to a H atom in ground state so that it can emit an

H_{γ}

line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such

H_{γ}

photon?

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>LA>Q 12.24

The first four spectral lines in the Lyman series of a H-atom are

λ = 1218 \overset{\circ}{A,} 1028 \overset{\circ}{A}, 974.3 \overset{\circ}{A} and 951.4 \overset{\circ}{A} .

If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>LA>Q 12.25

Deutrium was discovered in

1932

by Harold Urey by measuring the small change in wavelength for a particular transition in

^{1} H

and

^{2} H

. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass

μ

, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here

μ = m_{e} M / (m_{e} + M) w h e r e M

is the nuclear mass and

m_{e}

is the electronic mass. Estimate the percentage difference in wavelength for the first line of Lyman series in

{}^{1}H and^{2} H

(Mass of^{1} H nucleus is 1.6725 \times 10^{- 27} kg, Mass^{2} H nucleus is 3.3374 \times 10^{- 27} kg, Mass of electron = 9.109 \times 10^{- 31} kg .)

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>LA>Q 12.26

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $H$ -atom when $R = 0.1 Å$

MEDIUM

12th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 12>Chapter 12 - Atoms>LA>Q 12.26

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $H$ -atom when $R = 10 Å$

Use, Ground state radius $r_{A} = \frac{h^{2}}{4 π^{2} m K e^{2}} = 0.53 Å$ , $P . E . = \frac{e^{2}}{4 π ε_{0} r_{A}} = 27.2 e V$

Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Important Questions on Atoms

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Positronium is just like a $H$ -atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?