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. $\frac{6.6 \times 10^{–9}}{\sqrt{1664}}$
• D. $\frac{6.6 \times 10^{–9}}{\sqrt{166.4}}m$

Answer 1

Answer: (B)

$\frac{6.6 \times 10^{–9}}{\sqrt{145.6}}m$

Answer 2

K.E of e^– ejected = Energy of incident quantum – threshold

energy

= 5 eV

l = $\frac{h}{mV} = \frac{h}{\sqrt{2mK.E}}$

= $\frac{6.6 \times 10^{–34}}{\sqrt{2 \times 9.1 \times 10^{–31} \times 5 \times 1.6 \times 10^{–19}}}$m = $\frac{6.6}{\sqrt{145.6}}$m