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VITEEE 2018 Paper

Section: Physics

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Question : 6 of 125

Marks: +1, -0

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength

\lambda_1

\lambda_2

emitted in the two cases is

7/5
27/20
27/5
20/7

Solution:

Here, for wavelength

\lambda_1

n_1=4

and

n_2=3

and for

\lambda_2, n_1=3

and

n_2=2

We have

\frac{hc}{\lambda} = -13.6 \left[ \frac{1}{n_2^2} - \frac{1}{n_1^2} \right]

so, for

\lambda_1

\Rightarrow \frac{hc}{\lambda_1} = -13.6 \left[ \frac{1}{4^2} - \frac{1}{3^2} \right]

\frac{hc}{\lambda_1} = -13.6 \left[ \frac{7}{144} \right] \dots

(i)

simiarly, for

\lambda_2

\Rightarrow \frac{hc}{\lambda_1} = -13.6 \left[ \frac{1}{3^2} - \frac{1}{2^2} \right]

\frac{hc}{\lambda_1} = -13.6 \left[ \frac{5}{36} \right] \dots

(ii)

Hence, from Eqs. (i) and (ii), we get

\therefore \frac{\lambda_1}{\lambda_2} = \frac{-13.6 \left[ \frac{7}{144} \right]}{-13.6 \left[ \frac{5}{36} \right]} = \frac{20}{7}

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