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The wavelength of radiation emitted is\mathrm{\lambda_0} when an electron jumps from the third to second orbit of hydrogen atom. For the electron jump from fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be
 

Option: 1

(16/25) \lambda_0
 


Option: 2

(20/27) \lambda_0
 


Option: 3

(27/20) \lambda_0
 


Option: 4

(25/16) \lambda_0


Answers (1)

best_answer

\mathrm{\text {} \frac{\lambda}{\lambda_0}=\frac{\left[\frac{1}{2^2}-\frac{1}{3^2}\right]}{\left[\frac{1}{2^2}-\frac{1}{4^2}\right]}=\frac{5}{36} \times \frac{16}{3}=\frac{20}{27} \Rightarrow \lambda=\frac{20}{27} \lambda_0}

Hence option 2 is correct.

Posted by

Ritika Jonwal

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