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Q.48) The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $\mathrm{n}-2 \rightarrow \mathrm{n}=3$ and $\mathrm{n}=4 \rightarrow \mathrm{n}=6$ transitions, respectively, is

A) $\frac{1}{4}$ 

B) $\frac{1}{36}$

C) $\frac{1}{16}$

D) $\frac{1}{9}$

 

Answers (1)

best_answer

 Given that,

Transition A: $n=2 \rightarrow n=3$
Transition B: $n=4 \rightarrow n=6$

We have 

$$
\Delta E \propto\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$$


And since $\lambda \propto \frac{1}{\Delta E}$, we can write:

$$
\frac{\lambda_1}{\lambda_2}=\frac{\Delta E_2}{\Delta E_1}
$$

For $n=2 \rightarrow 3$ :

$$
\Delta E_1 \propto\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}
$$


For $n=4 \rightarrow 6$ :

$$
\Delta E_2 \propto\left(\frac{1}{4^2}-\frac{1}{6^2}\right)=\left(\frac{1}{16}-\frac{1}{36}\right)=\frac{5}{144}
$$
so the ratio is $\frac{\lambda_1}{\lambda_2}=\frac{\Delta E_2}{\Delta E_1}=\frac{\frac{5}{144}}{\frac{5}{36}}=\frac{1}{4}$

Hence the correct option is (1)

Posted by

Saumya Singh

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