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Question

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


Solution

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

Physics

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