Q. 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 photo electrons ? $(c = 3 \times 10^{8} ms^{-1}, h = 6.6 \times 10^{-34} J-s)$

Solution:

$B = B_0 \sin (\pi \times 10^7C)t + B_0 \sin (2 \pi \times 10^7C)t$ 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 \times10^{7} C\right)t \,v_{1} = \frac{10^{7}}{2} \times C $
$ B_{2} = B_{0} \sin \left(2 \pi \times10^{7} C\right)t \,v_{2} = 10^{7} C $
where C is speed of light $ C =3 \times10^{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
$v = \phi + KE_{max}$
energy of photon
$E_{ph} = h_{V} = 6.6 \times10^{-34} \times10^{7} \times3 \times10^{9} $
$ E_{ph} = 6.6 \times3 \times10^{-19 }J $
$ = \frac{6.6 \times3 \times10^{-19}}{1.6 \times10^{-19}} eV = 12.75 eV$
$ KE_{max } = E_{ph} - \phi $
= 12.375 - 4.7 = 7.675 eV $\approx$ 7.7 eV

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