Question: 18: When two monochromatic lights of frequency, v and \frac{v}{2} are incident on a photoelectric metal, their stopping potential becomes V_{s} and \frac{V_{s}}{2} respectively. The threshold frequency for this metal is
(1) \frac{2}{3} v
(2) \frac{3}{2} v
(3) 2 v
(4) 3 v
Answer: Option (2)
Explanation:
According to Einstein’s photoelectric equation, the stopping potential is related to the frequency of incident light by
eV_s = h(v - v_0),
where v is the frequency of incident light and v_0
is the threshold frequency of the metal.
For incident light of frequency v, the stopping potential is V_s.
Hence, eV_s = h(v - v_0).
For incident light of frequency \frac{v}{2},
the stopping potential is \frac{V_s}{2}.
Hence, e\frac{V_s}{2} = h\left(\frac{v}{2} - v_0\right).
Dividing the second equation by the first,
\frac{1}{2} = \frac{\frac{v}{2} - v_0}{v - v_0}.
Cross multiplying,
v - v_0 = v - 2v_0.
v_0 = \frac{v}{2}.
Substituting v_0 into the first equation,
we get the threshold frequency in terms of the given frequency as
v_0 = \frac{3}{2}v.
Hence, the threshold frequency of the metal is \frac{3}{2}v,
and the correct answer is option (2).