Discussion Forum



1)

A certain  metal when irradiated by light (r=3.2 x 1016 Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (r=2.0 x 1016 Hz)

The v0 of metal is 


A) $1.2 \times 10^{14}Hz$

B) $8 \times 10^{15}Hz$

C) $1.2 \times 10^{16}Hz$

D) $4 \times 10^{12}Hz$

Answer:

Option B

Explanation:

(kE)1 = hv1- hv0

 (kE)2= hv2-hv0

 As, (KE)1 =2 x (KE)2

$\therefore$    $hv_{1}-hv_{0}=2(hv_{2}-hv_{0})$

or    $hv_{0}=2hv_{2}-hv_{1}$

 or   $v_{0}=2v_{2}-v_{1}$

 $=2\times (2\times 10^{16})-(3.2\times 10^{16})$

 $=0.8\times 10^{16}Hz=8\times 10^{15}Hz$