Question

When the electron orbiting in hydrogen atom in its ground state mo ves to the third excited state, show how the de broglie wavelength associated with it would be affected?

Answer 1

Wavelength will increase 4 times

Answer 2

De-Broglie wavelength of electron:

The De-Broglie wavelength when the particle is in ground state is given by the formula:

$\lambda_{0}=\frac{h}{\sqrt{2 m E_{0}}}$

The De-Broglie wavelength when the particle is in third excited state is given by the formula:

$\lambda_{3}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{3}}}$

Now, on dividing the wavelength at ground state and third excited state, we get,

$\frac{\lambda_{0}}{\lambda_{3}}=\sqrt{\frac{E_{3}}{E_{0}}}$

$\frac{\lambda_{0}}{\lambda_{3}} \propto \sqrt{\frac{E_{3}}{E_{0}}}$

Wavelength and energy are proportional to each other.

Thus, when the electron jumps from ground state to third excited state, it affects the energy of the electron.