The threshold frequency \( u_0 \) for a metal is \( 7.0 \times 10^{14} \text{ s}^{-1} \). Calculate the kinetic energy of an electron emitted when radiation of frequency \( \nu = 1.0 \times 10^{15} \text{ s}^{-1} \) hits the metal.

According to Einstein's photoelectric equation, the kinetic energy (\( K.E. \)) of the emitted electron is given by: \( K.E. = h(\nu - \nu_0) \). Given: Planck's constant \( h = 6.626 \times 10^{-34} \text{ J s} \), Incident frequency \( \nu = 1.0 \times 10^{15} \text{ s}^{-1} = 10 \times 10^{14} \text{ s}^{-1} \), Threshold frequency \( \nu_0 = 7.0 \times 10^{14} \text{ s}^{-1} \). Substituting the values: \( K.E. = 6.626 \times 10^{-34} \times (10 - 7) \times 10^{14} = 6.626 \times 10^{-34} imes 3.0 \times 10^{14} = 1.988 \times 10^{-19} \text{ J} \).