Question

A certain metal when irradiated by light $(r = 3.2 \times 10^{16} Hz)$ emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light $(r =2.0 \times 10^{16} Hz)$. The $v_0$ 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 1

Answer: (B)

$(KE)_1 = hv_1 - hv_0$
$(KE)_2 = hv_2 - hv_0$
As, $(KE)_1 = 2 \times (KE)_2$
$ \therefore \, \, hv_1 - hv_0 = 2(hv_2 - hv_0)$
or, $hv_0 = 2hv_2 - hv_1$
or, $v_0 = 2v_2 - v_1$
$= 2 \times (2 \times 10^{16}) - (3.2 \times 10^{16})$
$= 0.8 \times 10^{16} \, Hz = 8 \times 10^{15} \, Hz$