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Atoms and Nuclei Part 3

Section: Physics

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Question : 41 of 98

Marks: +1, -0

During the transition of electron from state

A

C

of a Bohr atom, the wavelength of emitted radiation is

2000\text{Å}

6000\text{Å}

when the electron jumps from state

B

C

. Then the wavelength of the radiation emitted during the transition of electrons from state

A

B

[24 Jan 2025 Shift 1]

$2000\text{Å}$
$3000\text{Å}$
$6000\text{Å}$
$4000\text{Å}$

Solution:

For

A

to

C

\frac{h c}{\lambda_1}=E_0 z^2 \left(\frac{1}{n_c^2}-\frac{1}{n_A^2}\right)

\text{ And } \frac{h c}{\lambda_2}=E_0 z^2 \left(\frac{1}{n_c^2}-\frac{1}{n_B^2}\right)

So for

A

and

B

\frac{h c}{\lambda_3}=E_0 z^2 \left(\frac{1}{n_B^2}-\frac{1}{n_A^2}\right)

Clearly subtracting equation (ii) from equation (i)

h c \left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right] = E_0 z^2 \left[\frac{1}{n_B^2}-\frac{1}{n_A^2}\right] = \frac{h c}{\lambda_3}

\Rightarrow \frac{1}{\lambda_3} = \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \Rightarrow \frac{1}{\lambda_3} = \frac{6000-2000}{6000 \times 2000} = \frac{1}{3000}

\lambda_3=3000\text{Å}

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