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If photons of frequency v are incident on the surfaces of metals. A & B of threshold frequencies \frac{v}{2} and \frac{v}{3} respectively, the ratio of the maximum kinetic energy of electrons emitted from A to that from B is
Option: 1 2 : 3
Option: 2 3 : 4
Option: 3 1 :3
Option: 4 \sqrt{3}:\sqrt{2}

Answers (1)

\\\frac{KE_{A}}{KE_{B}}=\frac{h\nu-h\frac{\nu}{2}}{h\nu-h\frac{\nu}{3}}=\frac{3}{4}

 

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