A photon of wavelength 3000 \(\overset{^\circ }{\mathrm{A}}\) strikes a metal surface. The work function of the metal is…

Using photoelectric equation:

\(\mathrm{KE}=\mathrm{hν}−\mathrm{ϕ}\)

With \(\nu =\frac{\mathrm{c}}{\lambda }\)

\(\mathrm{E}=\frac{(6.626\times {10}^{−34})(3\times {10}^{8})}{3000\times {10}^{−10}}\)

= 6.6 × 10^–19 J

1eV = 1.6 × 10^–19 J

Therefore

\(\mathrm{E}=\frac{6.6\times {10}^{–19}}{1.6\times {10}^{–19}}=4.125\mathrm{eV}\)

\(\mathrm{KE}=4.125−2.13=1.99\approx 2.0\text{ eV}\)