Question

The magnetic field associated with a light wave is given, at the origin, by
$B=B_0[sin(3.14\times {10}^7)ct+sin(6.28\times {10}^7)ct].$ If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the electrons?
$(C=3\times {10}^{8}m{s}^{-1},h=6.6\times {10}^{-34}J-s)$

• A. 12.5 eV
• B. 8.52 eV
• C. 7.72 eV
• D. 6.82 eV

Answer 1

The correct option is C 7.72 eV

$\text{Given}$ : Magnetic field

$\text{To find}$ : maximum kinetic energy of the electrons

$\text{Solution}$ :

$B=B_0\sin (\pi \times {10}^{7}C)t+B_0\sin (2\pi \times {10}^{7}C)t$

since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency, therefore

$B_1=B_0\sin (\pi \times {10}^{7}C)t{v}_1=\frac{{10}^7}{2}\times C$

$B_2=B_0\sin (2\pi \times {10}^{7}C)t{v}_2={10}^{7}C$

where C is speed of light

Also, $C=3\times {10}^{8}m/s$

Therefore, $v_2>v_1$

so KE of photoelectron will be maximum for photon of higher energy.

$v_2={10}^{7}C$ Hz $v=\varphi +K{E}_{max}$

energy of photon, $E_{ph}=h_V=6.6\times {10}^{-34}\times {10}^7\times 3\times {10}^9$

$E_{ph}=6.6\times 3\times {10}^{-19}J$ $=\frac{6.6\times 3\times {10}^{-19}}{1.6\times {10}^{-19}}eV=12.75eV$ $K{E}_{max}=E_{ph}-\varphi$ = 12.375 - 4.7 = 7.675 eV $\approx$ 7.7 eV

$\text{Hence C is the correct option}$