Modern Physics 2 Question 4
4. The electric field of light wave is given as $\mathbf{E}=10^{-3} \cos \frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t \quad \hat{\mathbf{x}} NC^{-1}$. 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: 4. (a)
| 5. (c) | 6. (d) | 7. (a) | 8. (a) |
|---|---|---|---|
| 9. (c) | 10. (c) | 11. (b) | 12. (a) |
| 13. (b) | 14. (a) | 15. (b) | 16. (a) |
| 17. (a) | 18. (c) | 19. (b) | 20. (a, c) |
| 21. (a, c) | 22. (a, b, c) | 23. (b, d) | 24. (a, b, c) |
| 25. (c, d) | $\mathbf{2 6 .}$. | $\mathbf{2 7 . 7}$ | 28. Frequency |
| 29. Question is incomplete | $\mathbf{3 0 .} F$ | ||
| 32. (a) $5 \times 10^{7}$ (b) $2 \times 10^{3} N / C$ (c) $23 eV$ |
Solution:
- Given, $\mathbf{E}=10^{-3} \cos \frac{2 \pi x}{5 \times 10^{-7}}-2 \pi \times 6 \times 10^{14} t \quad \hat{\mathbf{x}} NC^{-1}$
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, $k=2 \pi / \lambda$ )
$$ \Rightarrow \quad \lambda=5 \times 10^{-7} m=5000 \AA $$
Or
The value of $\lambda$ can also be calculated as, after comparing the given equation of $\mathbf{E}$ with standard equation, we get
$$ \begin{array}{lll} & & \omega=6 \times 10^{14} \times 2 \pi \\ \Rightarrow & \nu=6 \times 10^{14} \quad[\because 2 \pi \nu=\omega] \\ \text { As, } & c=\nu \lambda \\ \Rightarrow & \lambda=\frac{c}{\nu}=\frac{3 \times 10^{8}}{6 \times 10^{14}}=5 \times 10^{-7} m=5000 \AA \end{array} $$
According to Einstein’s equation for photoelectric effect, i.e.,
$$ \frac{\dot{h} c}{\lambda}-\varphi=(KE) _{\max }=e V _0 $$
For photon, substituting the given values,
$$ \begin{aligned} E=\frac{h c}{\lambda} & =\frac{12375 eV}{\lambda} \\ \frac{h c}{\lambda} & =\frac{12375}{5000} eV \quad \text { [using Eq. (i)]. } \end{aligned} $$
Now, substituting the values from Eq. (iii) in Eq. (ii), we get
$$ \begin{array}{ll} & \frac{12375}{5000} eV-2 eV=e V _0 \\ \Rightarrow & 2.475 eV-2 eV=e V _0 \\ \text { or } & V _0=2.475 V-2 V \\ & =0.475 V \Rightarrow V _0 \approx 0.48 V \end{array} $$
