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)

$(KE{)}_1=h{v}_1-h{v}_0$
$(KE{)}_2=h{v}_2-h{v}_0$
As, $(KE{)}_1=2\times (KE{)}_2$
$\therefore h{v}_1-h{v}_0=2(h{v}_2-h{v}_0)$
or, $h{v}_0=2h{v}_2-h{v}_1$
or, $v_0=2v_2-v_1$
$=2\times (2\times 10^{16})-(3.2\times 10^{16})$
$=0.8\times 10^{16}Hz=8\times 10^{15}Hz$