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

We know De Broglie wavelength can be related as

$\lambda =\frac{h}{p}$ where h=planks constant and p= momentum of the electron.

Also momentum $p=mv$ and kinetic energy$E_K=\frac{1}{2}m{v}^2$

From above equation we can relate momentum p and kinetic energy as,

$p=\sqrt{2m{E}_K}$

Now ratio wavelength of hydrogen atom when the atom jumps from third excited state to ground state viz(n=4 to n=1)

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{h}{p_1}}{\frac{h}{p_2}}=\frac{p_2}{p_1}=\frac{\sqrt{2m{E}_{K2}}}{\sqrt{2m{E}_{K1}}}$ -------(A)

Also we know $E_{Kn}=\frac{-13.6Z^2}{n^2}$

so, ratio of wavelength of Ground state{n = 1 }nto 3rd excited state { n = 4}

Hence, answer is 1 : 4