Question

When the electron in a hydrogen atom jumps from second orbit to first orbit the wavelength of radiation emitted is lambda .when electron jumps from third orbit to first the wavelength of emitted radiation would be?

Answer 1

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Answer 2

Answer: $\lambda'=\frac{27\lambda}{32}$

Explanation: $\frac{1}{\lambda}=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where, $\lambda$ = wavelength

Z = Atomic number = 1

$n_1$ = Lower energy level = 1

$n_2$ = Upper energy level = 2

$R_h$= Rydberg's constant

Putting values in above equation, we get:

$\frac{1}{\lambda}=R_H\times 1^2\left(\frac{1}{2^2}-\frac{1}{1^2}\right)$

$R_H=\frac{4}{3\lambda}$

$\frac{1}{\lambda'}=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

2)

$n_1$ = Lower energy level = 1

$n_2$ = Upper energy level = 3

$\frac{1}{\lambda'}=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\frac{1}{\lambda'}=\frac{4}{3\lambda}\times 1^2\left(\frac{1}{1^2}-\frac{1}{3^2}\right)$

$\lambda'=\frac{27\lambda}{32}$