Concept:Using Einstein's photoelectric equation to relate maximum kinetic energy, incident frequency, and threshold frequency.
Explanation:For a metal with threshold frequency
v0, the maximum kinetic energy is:
Kmax=h(v−v0)For frequency
v1, we have:
K1=h(v1−v0)For frequency
v2, we have:
K2=h(v2−v0)Given the ratio of maximum kinetic energies is
k:1:
K2K1=kSubstituting the expressions:
v2−v0v1−v0=kNow solve for
v0:
v1−v0=k(v2−v0)v1−v0=kv2−kv0Rearranging terms:
kv0−v0=kv2−v1(k−1)v0=kv2−v1Therefore:
v0=k−1kv2−v1Answer:The threshold frequency of the metallic surface is
k−1kv2−v1. Hence, the correct option is D.