Question

Given the ground state energy $E_0=-13.6eV$ and Bohr radius $a_0$ = 0.53 A . Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.

Answer 1

Ground state energy is $E_o=-13.6eV$

Energy in the first excited state=$E_1=\frac{-13.6eV}{(2{)}^2}=-3.4eV$ We know that, the de Broglie wavelength λ is given as $lambda$

$\lambda =\frac{h}{p}But,p=\sqrt{2m{E}_1}\therefore \lambda =\frac{h}{\sqrt{2m{E}_1}}=\frac{6.63\times {10}^{-34}}{\sqrt{2\times (9.1\times {10}^{-31})\times }(3.4\times 1.6\times {10}^{-19})}=\frac{6.63\times {10}^{-}34}{9.95\times {10}^{-25}}\lambda =6.6\times {10}^{-10}m$