Question

Suppose an electron in a hydrogen atom in the ground state absorbs 1.75 times as much energy as the minimum required for it to escape from the atom. What is the wavelength of the emitted electron?

Answer 1

SOLUTION : Since $13.6 cV$ is needed for complete removal of electron, the energy absorbed $=13.6 \times 1.75 cV =23.8 cV$. The excess energy $(23.8-13 \cdot 6) cV =10 \cdot 2 eV$ is converted to $KE$ of the emitted electron.

Now $10.2 eV =10.2 \times 1.60 \times 10^{-19} J =1.632 \times 10^{-18} J$

$K E=\frac{1}{2} m v^{2}, v=\left(\frac{2 K E}{m}\right)^{\frac{1}{2}} \text { or, } v=\frac{\left(2 \times 1.632 \times 10^{-18}\right)^{1 / 2}}{\left(9 \cdot 1 \times 10^{-31}\right)^{1 / 2}}=1 \cdot 894 \times 10^{6} ms ^{-1}$

According to de Broglie relation

$\lambda=\frac{h}{m v}=\frac{6 \cdot 626 \times 10^{-34}}{9 \cdot 1 \times 10^{-31} \times 1 \cdot 894 \times 10^{6}} m =3 \cdot 84 \times 10^{-10} m .$