Modern Physics Ques 101
- The electric field of light wave is given as $\mathbf{E}=10^{-3} \cos (\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t) \hat{\mathbf{x}} NC^{-1}$. This light falls on a metal plate of work function $2 $ $eV$. The stopping potential of the photoelectrons is
Given, $E($ in $eV)=\frac{12375}{\lambda(\text { in } \AA)}$
(Main 2019, 9 April I)
(a) $0.48 $ $V$
(b) $0.72 $ $V$
(c) $2.0 $ $V$
(d) $2.48 $ $V$
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Answer:
Correct Answer: 101.(a)
Solution:
Formula:
- Given, $\mathbf{E}=10^{-3} \cos \left(\frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t \right) \hat{\mathbf{x}} NC^{-1}$
By comparing it with the general equation of electric field of light, i.e.
$ \begin{aligned} & E=E _0 \cos (k x-\omega t) \hat{\mathbf{x}} \text {, we get } \\ & k=\frac{2 \pi}{5 \times 10^{-7}}=2 \pi / \lambda \end{aligned} $
(from definition, $k=2 \pi / \lambda$ )
$ \Rightarrow \lambda=5 \times 10^{-7} m=5000 \AA $ $\quad$ …….(i)
The value of $\lambda$ can also be calculated as, after comparing the given equation of $\mathbf{E}$ with standard equation, we get
$ \omega=6 \times 10^{14} \times 2 \pi $
$\Rightarrow \nu=6 \times 10^{14} \quad[\because 2 \pi \nu=\omega] $
$\text { As, } c=\nu \lambda $
$\Rightarrow \lambda=\frac{c}{\nu}=\frac{3 \times 10^{8}}{6 \times 10^{14}}=5 \times 10^{-7} m=5000 \AA$
According to Einstein’s equation for photoelectric effect, i.e.,
$ \frac{\dot{h} c}{\lambda}-\phi=(KE) _{\max }=e V _0 $$\quad$ …….(ii)
For photon, substituting the given values,
$E=\frac{h c}{\lambda} =\frac{12375 eV}{\lambda} $
$\frac{h c}{\lambda} =\frac{12375}{5000} eV \quad \text { [using Eq. (i)]. } $ $\quad$ …….(iii)
Now, substituting the values from Eq. (iii) in Eq. (ii), we get
$ \frac{12375}{5000} eV-2 eV=e V _0 $
$\Rightarrow 2.475 eV-2 eV=e V _0 $
$\text { or } V _0=2.475 V-2 V $
$ =0.475 V \Rightarrow V _0 \approx 0.48 V$
BETA
