For electron

De – Broglie wavelength, \( \lambda _{c} = \frac{h}{p} \)

Where:

p is momentum, p = mv

E is the energy, E =\( \frac{1}{2}mv^{2} \)

E =\( \frac{1}{2}mv^{2} \) \( \Rightarrow \frac{1}{2} \frac{p^{2}}{m} \) \( \Rightarrow p = \sqrt{2mE} \) \( \Rightarrow \lambda _{c} = \frac{h}{\sqrt{2mE}} \)

Therefore, the De – Broglie wavelength, \( \lambda _{c} = \frac{h}{\sqrt{2mE}} \)

For photon energy, E =\( \frac{hc}{\lambda} \) \( \Rightarrow \lambda = \frac{hc}{E} \) \( \Rightarrow \frac{\lambda _{c}}{\lambda} = \frac{h}{\sqrt{2mE}} \frac{E}{hc} \) \( \Rightarrow \frac{1}{C}\sqrt{\frac{E}{2m}} \)

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