Question

On illuminating a metal surface by $$ 8.5 \times 10^{14} Hz$$, the emitted-electron's maximum kinetic energy is $$0.52\,eV $$. For the same surface on illuminating with light of frequency $$ 12.0 \times 10^{14}\,Hz$$, the maximum kinetic energy of photoelectrons is $$ 1.97\,eV$$. Find out the work function of the metal.

Answer 1

$$ hv = \dfrac{1}{2} mv^2 + \phi $$

$$ 6.63 \times 10^{-34} \times 8.5 \times 10^{14} = 0.52 \,eV + \phi $$

$$ 0.52 \,eV + \phi = \dfrac{6.63 \times 10^{-34} \times 8.5 \times 10^{14}}{1.6 \times 10^{-19}}\,eV $$

$$ = 3.52\,eV$$

$$ \therefore $$ $$ \phi = (3.52 - 0.52)\,eV = 3\,eV $$