When two monochromatic lights of frequency, \( \nu \) and \( 2\nu \) are incident on a photoelectric metal, their stopping potential becomes \( \frac{V_s}{2} \) and \( V_s \) respectively. The threshold frequency for this metal is:

Using Einstein's photoelectric equation: \( eV_s = h\nu - h\nu_0 \). For frequency \( \nu \), \( e\frac{V_s}{2} = h\nu - h\nu_0 \). For frequency \( 2\nu \), \( eV_s = h(2\nu) - h\nu_0 \). Dividing the two gives \( \frac{1}{2} = \frac{\nu - \nu_0}{2\nu - \nu_0} \Rightarrow 2\nu - \nu_0 = 2\nu - 2\nu_0 \) which simplifies to \( \nu_0 = \frac{2}{3}\nu \).