Question

When a certain metal was irradiated with light of frequency $3.2\times {10}^{16}Hz$, the photoelectrons emitted has twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency $2.0\times {10}^{16}Hz$. Calculate $v_0$ for the metal.

Answer 1

Applying photoelectric equation,
$KE=hv-h{v}_0$
or $(v-v_0)=\frac{KE}{h}$

Given, ${KE}_2=2{KE}_1$

$v_2-v_0=\frac{{KE}_2}{h}$ ...(i)
and $v_1-v_0=\frac{{KE}_1}{h}$ ...(ii)

Dividing equation (i) by equation (ii),
$\frac{v_2-v_0}{v_1-v_0}=\frac{{KE}_2}{{KE}_1}=\frac{2{KE}_1}{{KE}_1}=2$

or $v_2-v_0=2v_1-2v_0$

or $v_0=2v_1-v_2=2(2.0\times {10}^{16})-(3.2\times {10}^{16})$

$=8.0\times {10}^{15}Hz$.