Atomic StructureHard
Question
Photoelectric emission is observed from a metal surface for frequencies v1 and v2 of the incident radiation (v1 > v2). If maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1: K, then the threshold frequency for the metal is given by
Options
A.$\frac{v_{2} - v_{1}}{K - 1}$
B.$\frac{Kv_{2} - v_{1}}{K - 1}$
C.$\frac{Kv_{1} - v_{2}}{K}$
D.$\frac{Kv_{1} - v_{2}}{K - 1}$
Solution
$h\upsilon_{1} = h\upsilon_{0} + E\text{ and }h\upsilon_{2} = h\upsilon_{0} + E.K$
$\therefore\upsilon_{0} = \frac{K\upsilon_{1} - \upsilon_{2}}{K - 1}$
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