Modern Physics 2 Question 8
8. The magnetic field associated with a light wave is given at the origin, by
$B=B _0\left[\sin \left(3.14 \times 10^{7}\right) c t+\sin \left(6.28 \times 10^{7}\right) c t\right]$.
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 photoelectrons?
(Main 2019, 10 Jan II) (Take, $c=3 \times 10^{8} ms^{-1}$ and $h=6.6 \times 10^{-34} J$-s)
(a) $7.72 eV$
(b) $6.82 eV$
(c) $8.52 eV$
(d) $12.5 eV$
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Solution:
- According to question, the wave equation of the magnetic field which produce photoelectric effect
$$ B=B _0\left[\left(\sin \left(3.14 \times 10^{7} c t\right)+\sin \left(6.28 \times 10^{7} c t\right)\right]\right. $$
Here, the photoelectric effect produced by the angular frequency $(\omega)=6.28 \times 10^{7} c$
$$ \begin{aligned} \Rightarrow \quad \omega & =6.28 \times 10^{7} \times 3 \times 10^{8} \\ \omega & =2 \pi \times 10^{7} \times 3 \times 10^{8} rad / s \end{aligned} $$
Using Eqs. (i)
$$ \begin{aligned} & h \nu=\frac{h \omega}{2 \pi}=\frac{h \times 2 \pi \times 10^{7} \times 3 \times 10^{8}}{2 \pi} \\ & h \nu=12.4 eV \end{aligned} $$
Therefore, according to Einstein equation for photoelectric effect
$$ \begin{aligned} & E=h \nu=\varphi+KE _{\max } \\ & \Rightarrow \quad K E _{\max }=E-\varphi \\ & \text { (where, } \varphi=\text { work-function }=4.7 eV \text { ) } \\ & KE _{\max }=12.4-4.7=7.7 eV \\ & \text { or } \quad KE _{\max }=7.7 eV \end{aligned} $$
