Question

The magnetic field associated with a light wave is given, at the origin, by $B=B_0\left[sin\right(3.14\times {10}^7)ct+sin(6.28\times {10}^7\left)ct\right]$. 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

The correct option is A 7.72 eV

$B=B_{0}sin(\pi \times {10}^{7}C)t+B_{0}sin(2\pi \times {10}^{7}C)t$
since there are two EM waves with different
frequency, to get maximum kinetic energy we
take the photon with higher frequency
$B_1=B_{0}sin(\pi \times {10}^{7}C)t{v}_1=\frac{{10}^7}{2}C$
$B_2=B_{0}sin(2\pi \times {10}^{7}C)t{v}_2={10}^{7}C$
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.
$v_2={10}^{7}CHz$
$hv=\varphi +K{E}_{max}$
energy of photon
$E_{ph}=hv=6.6\times {10}^{-34}\times {10}^7\times 3\times {10}^9$
$e_{ph}=6.6\times 3\times {10}^{-19}J$
$=\frac{6.6\times 3\times {10}^{-19}}{1.6\times {10}^{-19}}eV=12.375eV$
$K{E}_{max}=E_{ph}-\varphi$
$=12.375-4.7=7.675eV\approx 7.72eV$