Question

The energy of an electron having de-Broglie wavelength ${\displaystyle \lambda }$ is ______.

(h = Plank's constant, m = mass of electron)

Options

• ${\displaystyle \frac{\text{h}}{2\text{m}\lambda }}$

• ${\displaystyle \frac{{\text{h}}^2}{2\text{m}{\lambda }^2}}$

• ${\displaystyle \frac{{\text{h}}^2}{2{\text{m}}^2{\lambda }^2}}$

• ${\displaystyle \frac{{\text{h}}^2}{2{\text{m}}^2\lambda }}$

Answer 1

The energy of an electron having de-Broglie wavelength ${\displaystyle \lambda }$ is ${\displaystyle \underline{\frac{{\text{h}}^2}{2\text{m}{\lambda }^2}}}$.

(h = Plank's constant, m = mass of electron)

Explanation:

de-Broglie wavelength

λ = ${\displaystyle \frac{\text{h}}{\text{p}}\frac{\text{h}}{\sqrt{2\text{m K.E}}}}$

or, λ² = ${\displaystyle \frac{{\text{h}}^2}{2\text{m}(\text{K}.\text{E})}}$ 

∴ K. E. = ${\displaystyle \frac{{\text{h}}^2}{2\text{m}{\lambda }^2}}$