Question

A hydrogen atom initially in the ground state absorbs a photon and is in the excited state with energy 12.5ev. calculate the longest wavelength of the radiation emitted and identify the series to which it belongs

Answer 1

sorry no answer but can you do rhis

Answer 2

Answer: The longest wavelength will be $\lambda= 6.57\times10^{-7}\ m$ and the series will Lyman series.

Explanation:

Given that,

Energy E = 12.5 e V

Hydrogen atom jumps from ground state to excited state with energy 12.5 e V

For this energy

$E = 13.6(\dfrac{1}{n_{1}^2}-\dfrac{1}{n_{2}^2})$

$12.5=13.6(\dfrac{1}{1}-\dfrac{1}{n_{2}^2})$

$n_{2}= 3.5$

This transition is not allowed.

So, take $n_{2}= 3$

Now, longest wave length will be

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_{3}^2}-\dfrac{1}{n_{2}^2})$

$\dfrac{1}{\lambda}=109587(\dfrac{1}{9}-\dfrac{1}{4})$

$\dfrac{1}{\lambda}= -15220.42$

$\lambda= -6.57\times10^{-7}\ m$

Negative sigh represent the loose of energy.

Hence, The longest wavelength will be $\lambda= 6.57\times10^{-7}\ m$ and the series will Lyman series.