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 $2.0eV.$ 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,
$\frac{1}{2}m{v}_{\max }^2-hv-w_0...(i)$
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)
$K{E}_{\max }=\left(6.63\times 10^{-34}\times 2\pi \times 10^{15}-2\times 1.6\times 10^{19}\right)+\hat{j}$
Hence $K{E}_{\max }=6.3e.V$