Modern Physics Ques 105
- 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?
(Take, $c=3 \times 10^{8} ms^{-1}$ and $h=6.6 \times 10^{-34} J$-s)
(Main 2019, 10 Jan II)
(a) $7.72 $ $eV$
(b) $6.82 $ $eV$
(c) $8.52 $ $eV$
(d) $12.5 $ $eV$
Show Answer
Answer:
Correct Answer: 105.(a)
Solution:
Formula:
- 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 \quad …….(i) \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=\phi+KE _{\max } \\ & \Rightarrow \quad K E _{\max }=E-\phi \\ & \text { (where, } \phi=\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} $
BETA
