Question

The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is 5eV. The de Broglie wavelength of the electron is -

• A. $\frac{6.6\times {10}^{-9}}{\sqrt{1456}}m$
• B. $\frac{6.6\times {10}^{-9}}{\sqrt{145.6}}m$
• C. $\frac{6.6\times {10}^{-9}}{\sqrt{1664}}m$
• D. $\frac{6.6\times {10}^{-9}}{\sqrt{166.4}}m$

Answer 1

The correct option is B $\frac{6.6\times {10}^{-9}}{\sqrt{145.6}}m$

K.E of $e^{-}$ ejected = Energy of incident quantum – threshold energy
= 5 eV
$\lambda =\frac{h}{mV}=\frac{h}{\sqrt{2mK.E}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 5\times 1.6\times {10}^{-19}}}m=\frac{6.6}{145.6}m$