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:

  1. 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

$$ \begin{array}{ll} & \frac{12375}{5000} eV-2 eV=e V _0 \\ \Rightarrow & 2.475 eV-2 eV=e V _0 \\ \text { or } & V _0=2.475 V-2 V \\ & =0.475 V \Rightarrow V _0 \approx 0.48 V \end{array} $$



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