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As per given figure A, B and C are the first, second and third excited energy levels of hydrogen atom
respectively. If the ratio of the two wavelengths  \left(\text { i.e. } \frac{\lambda_1}{\lambda_2}\right) \text { is } \frac{7}{4 n}  ,   then the value of n will be ______.

                                                      

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

For A, n = 2
B, n = 3
C, n = 4

\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\ & \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \\ & \frac{1}{\lambda_2}=\frac{7 \mathrm{R}}{144} \\ \end{aligned}  ......(1)

$$ \begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & \frac{1}{\lambda_1}=\frac{5 \mathrm{R}}{36} \end{aligned}    ......(2)

\begin{aligned} & \frac{\lambda_1}{\lambda_2}=\frac{7}{20}=\frac{7}{4 \times 5} \\ & \mathrm{n}=5 \end{aligned}

 

 

Posted by

Kuldeep Maurya

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