A metal has a work function of 3.0 eV. What is the threshold frequency of the metal when electrons are just liberated from the metal.(1 eV = 1.6 × 10-19J; h = 6.6 × 10-34Js)
4.8 × 10-19Hz
9.3 × 10-7Hz
4.8 × 1014Hz
7.3 × 1014Hz
9.3 × 1014Hz
Correct answer is D
The work function (\(\phi\)) of a metal is defined as:
\(\phi = hf_0\)
\(f_0 = \frac{\phi}{h}\)
\(\phi = 3.0 eV = 3.0 \times 1.6 \times 10^{-19}\)
= \(4.8 \times 10^{-19} J\)
\(f_0 = \frac{4.8 \times 10^{-19}}{6.6 \times 10^{34}}\)
= \(0.727 \times 10^{15} s^{-1}\)
= \(7.27 \times 10^{14} s^{-1}\)
\(\approxeq 7.3 \times 10^{14} Hz\)