Question

What amount of energy should be added to an electron to reduce its de Broglie wavelength from 100 to 50 pm?

Answer 1

Momentum of the particle $p=\frac{h}{\lambda }$
Initial momentum $p_i=\frac{6.63\times {10}^{-34}}{100\times {10}^{-12}}=6.63\times {10}^{-24}kg.m/s$
Initial energy $E_i=\frac{p_i^2}{2m}=\frac{(6.63\times {10}^{-24}{)}^2}{2\times 9.1\times {10}^{-31}}=2.4\times {10}^{-17}J$
Final momentum $p_f=\frac{6.63\times {10}^{-34}}{50\times {10}^{-12}}=13.26\times {10}^{-24}kg.m/s$
Final energy $E_f=\frac{p_f^2}{2m}=\frac{(13.26\times {10}^{-24}{)}^2}{2\times 9.1\times {10}^{-31}}=9.7\times {10}^{-17}J$
Thus energy added $\Delta E=E_f-E_i=(9.7-2.4)\times {10}^{-17}=7.3\times {10}^{-17}J$
$⟹\Delta E=\frac{7.3\times {10}^{-17}}{1.6\times {10}^{-19}}eV=450eV$