Question

The de Broglie wavelength of an electron accelerated to a potential of $400V$ is approximately

• A. $0.03nm$
• B. $0.04nm$
• C. $0.12nm$
• D. $0.06nm$

Answer 1

The correct option is D $0.06nm$

Electron is accelerated to a potential of $400V$.

Thus its kinetic energy $K=400eV$

de-Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$

$\therefore$ $\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2(9.1\times {10}^{-31})\times 400\times 1.6\times {10}^{-19}}}$

Or $\lambda =\frac{6.6\times {10}^{-34}}{1.08\times {10}^{-23}}=6.1\times {10}^{-11}$ $m$ $⟹$ $\lambda =0.06nm$