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 $4.7eV$, what will be the maximum kinetic energy of the photo electrons ? $(c=3\times 10^{8}m{s}^{-1},h=6.6\times 10^{-34}J-s)$

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

Answer 1

Answer: (A)

$B=B_0\sin (\pi \times 10^{7}C)t+B_0\sin (2\pi \times 10^{7}C)t$ since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
$B_1=B_0\sin \left(\pi \times 10^{7}C\right)t{v}_1=\frac{10^7}{2}\times C$
$B_2=B_0\sin \left(2\pi \times 10^{7}C\right)t{v}_2=10^{7}C$
where C is speed of light $C=3\times 10^8$ m/s
$v_2>v_1$
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
$E_{ph}=h_V=6.6\times 10^{-34}\times 10^7\times 3\times 10^9$
$E_{ph}=6.6\times 3\times 10^{-19}J$
$=\frac{6.6\times 3\times 10^{-19}}{1.6\times 10^{-19}}eV=12.75eV$
$K{E}_{max}=E_{ph}-\varphi$
$=12.375-4.7=7.675eV\approx 7.7eV$