Solveeit Logo
Login

Question

Question: Photoelectric emission is observed from a surface for frequencies n<sub>1</sub> and n<sub>2</sub> of...

Photoelectric emission is observed from a surface for frequencies n1 and n2 of incident radiations

(n1>n2). If the maximum K.E. of photoelectrons in two cases are in the ratio of 2 : 1, then threshold frequency n0 is given by -

A

ν2−ν12\frac{\nu_{2} - \nu_{1}}{2}

B

2ν1−ν2(2−1)\frac{2\nu_{1} - \nu_{2}}{(2 - 1)}

C

2ν2−ν1(2−1)\frac{2\nu_{2} - \nu_{1}}{(2 - 1)}

D

ν2−ν1\nu_{2} - \nu_{1}

Answer

2ν2−ν1(2−1)\frac{2\nu_{2} - \nu_{1}}{(2 - 1)}

Explanation

Solution

K.E1 = hn1 – hn0

K.E2 = hn2 – hn0

Ž 21=hν1−ν0hν2−hν0\frac{2}{1} = \frac{h\nu_{1} - \nu_{0}}{h\nu_{2} - h\nu_{0}}

Ž 2 hn2 – 2hn0 = hn1 – hn0

Ž n0 = 2ν2−ν11\frac{2\nu_{2} - \nu_{1}}{1}