Question

The electric field of certain radiation is given by the equation
$E = 200 \{ \sin (4\pi \times 10^{10}) t + \sin (4\pi \times 10^{15})t\}$
falls in a metal surface having work function $2.0\, eV.$ The maximum kinetic energy (in eV) of the photo electrons is
(use Planck’s constant $(h) = 6.63 \times 10^{-34} \, Js $ and electron charge $(E) = 1.6 \times 10^{-19 } C)$

• A. 3.3
• B. 4.3
• C. 5.3
• D. 6.3
• E. 7.3

Answer 1

Answer: (D)

$\because$ According Einstein's photoelectric equation,
$\frac{1}{2} m v_{\max }^{2}-h v-w_{0}\,...(i)$
Here, $w_{0}$ is work function and $v$ is frequency of photon.
According question equation, electric field repersenting given as, is
$E=200\left\{\sin \left(4 \pi \times 10^{10}\right) t+\sin \left(4 \pi \times 10^{15}\right) t\right\}$
Here, fundamental frequency of above equation will be LCM of both component frequency. Hence, the fundamental frequencyes
$\omega=4 \pi \times 10^{15} rad$
$\because \omega=2 \pi v$ or $h,\, v=\frac{4 \pi \times 10^{15}}{2 \pi}=2 \pi \times 10^{15}\, Hz$
Put the value of $h, v$ and $W_{0}$ in Eq. (i)
$KE _{\max }=\left(6.63 \times 10^{-34} \times 2 \pi \times 10^{15}-2 \times 1.6 \times 10^{19}\right)+\hat{ j }$
Hence $KE _{\max }=6.3\, e .V$