SATHEE: Modern Physics 2 Question 4

Modern Physics 2 Question 4

4. The electric field of light wave is given as $E = 10^{- 3} \cos \frac{2 π x}{5 \times 10^{- 7}} - 2 π \times 6 \times 10^{14} t \hat{x} N C^{- 1}$ . This light falls on a metal plate of work function $2 e V$ . The stopping potential of the photoelectrons is

Given, $E ($ in $e V) = \frac{12375}{λ (in \AA)}$

(Main 2019, 9 April I)

(a) $0.48 V$

(b) $0.72 V$

(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)	$26 .$ .	$27.7$	28. Frequency
29. Question is incomplete	$30 . F$
32. (a) $5 \times 10^{7}$ (b) $2 \times 10^{3} N / C$ (c) $23 e V$

Solution:

Given, $E = 10^{- 3} \cos \frac{2 π x}{5 \times 10^{- 7}} - 2 π \times 6 \times 10^{14} t \hat{x} N C^{- 1}$

By comparing it with the general equation of electric field of light, i.e.

$\begin{aligned} E = E_{0} \cos (k x - ω t) \hat{x}, we get \\ k = \frac{2 π}{5 \times 10^{- 7}} = 2 π / λ \end{aligned}$

(from definition, $k = 2 π / λ$ )

$\Rightarrow λ = 5 \times 10^{- 7} m = 5000 \AA$

Or

The value of $λ$ can also be calculated as, after comparing the given equation of $E$ with standard equation, we get

$\begin{array}{lll} ω = 6 \times 10^{14} \times 2 π \\ \Rightarrow & ν = 6 \times 10^{14} [∵ 2 π ν = ω] \\ As, & c = ν λ \\ \Rightarrow & λ = \frac{c}{ν} = \frac{3 \times 10^{8}}{6 \times 10^{14}} = 5 \times 10^{- 7} m = 5000 \AA \end{array}$

According to Einstein’s equation for photoelectric effect, i.e.,

$\frac{\dot{h} c}{λ} - φ = (K E)_{max} = e V_{0}$

For photon, substituting the given values,

$\begin{aligned} E = \frac{h c}{λ} & = \frac{12375 e V}{λ} \\ \frac{h c}{λ} & = \frac{12375}{5000} e V [using Eq. (i)]. \end{aligned}$

Now, substituting the values from Eq. (iii) in Eq. (ii), we get

$\begin{array}{ll} \frac{12375}{5000} e V - 2 e V = e V_{0} \\ \Rightarrow & 2.475 e V - 2 e V = e V_{0} \\ or & V_{0} = 2.475 V - 2 V \\ = 0.475 V \Rightarrow V_{0} \approx 0.48 V \end{array}$