Question

The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 eV) will be

• A. $4 \times 10^{10}Hz$
• B. $4 \times 10^{11}Hz$
• C. $4 \times 10^{14}Hz$
• D. $4 \times 10^{- 10}Hz$

Answer 1

Answer: (C)

$4 \times 10^{14}Hz$

Answer 2

Threshold wavelength $\lambda_{0} = \frac{12375}{W_{0}(eV)} = \frac{12375}{1.65} = 7500Å.$

∴ so minimum frequency

$\nu_{0} = \frac{c}{\lambda_{0}} = \frac{3 \times 10^{8}}{7500 \times 10^{- 10}} = 4 \times 10^{14}Hz.$