Threshold frequency,$v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0 \times 10^{15} \ s^{-1}$ was allowed to hit a metal surface,an electron having $1.988 \times 10^{-19} \ J$ of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to $600 \ nm$ hits the metal surface.

(N/A) We know that,$h v = h v_0 + KE$ or
$h v_0 = h v - KE = (6.626 \times 10^{-34} \ J s \times 1.0 \times 10^{15} \ s^{-1}) - 1.988 \times 10^{-19} \ J$
$h v_0 = 6.626 \times 10^{-19} \ J - 1.988 \times 10^{-19} \ J = 4.638 \times 10^{-19} \ J$
$v_0 = \frac{4.638 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J s} = 7.0 \times 10^{14} \ s^{-1}$
When,$\lambda = 600 \ nm = 600 \times 10^{-9} \ m$,the frequency of the incident photon is:
$v = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \ m s^{-1}}{600 \times 10^{-9} \ m} = 5.0 \times 10^{14} \ s^{-1}$
Since the incident frequency $v = 5.0 \times 10^{14} \ s^{-1}$ is less than the threshold frequency $v_0 = 7.0 \times 10^{14} \ s^{-1}$,no electron will be emitted.