Question

The magnetic field associated with a light waves given at the origin by $B=B_0\left[\sin \right(3.14\times {10}^7)\text{ct}+\sin (6.28\times {10}^7\left)\text{ct}\right]$
If this light falls on a silver plate having a work function of $4.7\text{eV}$, What will be the maximum kinetic energy of the photoelectrons?

($c=3\times {10}^8{\text{ms}}^{-1},h=6.6\times {10}^{-34}\text{J-s})$

• A. $\text{8.52 eV}$
• B. $\text{7.72 eV}$
• C. $\text{12.5 eV}$
• D. $\text{6.82 eV}$

Answer 1

The correct option is B $\text{7.72 eV}$

Accoriding to equation, there are two $EM$ waves with different frequency,

$B_1=B_0\sin (\pi \times {10}^{7}c)t$
and $B_2=B_0\sin (2\pi \times {10}^{7}c)t$
To get maximum kinetic energy we take the photon with higher frequency
using $B=B_0\sin \omega t$ and $\omega =2\pi \nu \Rightarrow \nu =\frac{\omega }{2\pi }$

$B_1=B_0\sin (\pi \times {10}^{7}c)t\Rightarrow {\nu }_1=\frac{{10}^7}{2}\times c$

$B_2=B_0\sin (2\pi \times {10}^{7}c)t\Rightarrow {\nu }_2={10}^{7}c$
where c is speed of light $c=3\times {10}^8\text{m/s}$
Clearly, ${\nu }_2>{\nu }_1$
So, $KE$ of photoelectrons will be maximum for photon of higher energy.
${\nu }_2={10}^{7}c\text{Hz}$
Using
$h\nu =\varphi +K{E}_{max}$
Energy of photon

$E_{ph}=h\nu =6.6\times {10}^{-34}\times {10}^7\times 3\times {10}^9$
$E_{ph}=6.6\times 3\times {10}^{-19}\text{J}$
$E_{ph}=\frac{6.6\times 3\times {10}^{-19}}{1.6\times {10}^{-19}}\text{eV}=12.375\text{eV}$

$K{E}_{max}=E_{ph}-\varphi =12.375-4.7=7.675\text{eV}\approx 7.7\text{eV}$