CBSE Class 12 · 2013 · Modern PhysicsLong Answer

WhatsApp Email

Using Bohr's postulates, derive the expression for the frequency of radiation emitted when the electron in a hydrogen atom undergoes a transition from a higher energy state (quantum number

n_{i}

) to a lower state (

n_{f}

). When the electron in a hydrogen atom jumps from energy state

n_{i} = 4

n_{f} = 3, 2, 1

, identify the spectral series to which the emission lines belong.

Answer: $ν = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3}} (1/ n_{f}^{2} - 1/ n_{i}^{2})$ ; $4 \to 3$ Paschen, $4 \to 2$ Balmer, $4 \to 1$ Lyman.

Solution

The energy of the

n^{th}

Bohr level is

E_{n} = - \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2} n ^{2}} .

When the electron falls from

n_{i}

n_{f}

, a photon of energy

h ν = E_{n_{i}} - E_{n_{f}}

is emitted:

h ν = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{2}} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) \Rightarrow ν = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3}} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) .

For

n_{i} = 4

4 \to 3

: Paschen series (infrared);

4 \to 2

: Balmer series (visible);

4 \to 1

: Lyman series (ultraviolet).

ν = \frac{m e ^{4}}{8 ε _{0}^{2} h ^{3}} (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})

CBSE Class 12 Board · 2013 · Subjective

Practise more: all CBSE board PYQs on Modern Physics →

CBSE Class 12 · 2013 · Modern PhysicsLong Answer

Solution

More board questions on Modern Physics