The threshold frequency $v_{0}$ for a metal is $7.0 \times 10^{14} \,s^{-1}.$ Calculate the kinetic energy of an electron emitted when radiation of frequency $v = 1.0 \times 10^{15} \,s^{-1}$ hits the metal.

According to Einstein's photoelectric equation:
Kinetic energy $(K.E.) = h(v - v_{0})$
Given:
$h = 6.626 \times 10^{-34} \,J \cdot s$
$v = 1.0 \times 10^{15} \,s^{-1} = 10.0 \times 10^{14} \,s^{-1}$
$v_{0} = 7.0 \times 10^{14} \,s^{-1}$
$K.E. = (6.626 \times 10^{-34} \,J \cdot s) \times (10.0 \times 10^{14} \,s^{-1} - 7.0 \times 10^{14} \,s^{-1})$
$K.E. = (6.626 \times 10^{-34} \,J \cdot s) \times (3.0 \times 10^{14} \,s^{-1})$
$K.E. = 1.9878 \times 10^{-19} \,J \approx 1.99 \times 10^{-19} \,J$