Question

When an electron in hydrogen atom jumps from the third excited state to the ground state. how would the de broglie wavelength have associated with the electron change?

Answer 1

$\bold{\text{we know,}}\\\\\lambda=\frac{h}{P}=\frac{h}{\sqrt{2mK.E}}\\\\so,\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{KE_2}{KE_1}}\\\\\text{\bold{we also know that}}\\KE_n=-13.6\frac{Z^2}{n^2}\\\\\text{put it above expression }},\\\frac{\lambda_1}{\lambda_2}=\frac{n_1}{n_2}$

so, ratio of wavelength of Ground state{n = 1 }nto 3rd excited state { n = 4}
$\bold{\frac{\lambda_{n=1}}{\lambda_{n=4}}}=\frac{1}{4}$

Hence, answer is 1 : 4