Question

which an electron in hydrogen atom jumps from the third excited state to the ground state

Answer 1

We know,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK.E}}$

So, $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{K{E}_2}{K{E}_1}}$

Also we know that

$K{E}_n=-13.6\frac{Z^2}{n^2}$

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}

$\frac{{\lambda }_{n=1}}{{\lambda }_{n=4}}=\frac{1}{4}$
Hence, answer is 1 : 4