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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