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======= ####39. The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} H z$ . If light of frequency $8.2 \times 10^{14} H z$ is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $2 V$

(b) $3 V$

(d) $1 V$

[2011M]

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Answer:

Correct Answer: 39. (a)

Solution:

(a) K.E. $= h v - h v_{th} = e V_{0} (V_{0} = .$ cut off voltage $)$

$\begin{aligned} \Rightarrow V_{0} = \frac{h}{e} (8.2 \times 10^{14} - 3.3 \times 10^{14}) \\ = \frac{6.6 \times 10^{- 34} \times 4.9 \times 10^{14}}{1.6 \times 10^{- 19}} \approx 2 V \end{aligned}$

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Dual Nature of Radiation and Matter - Result Question 37

39. The threshold frequency for a photosensitive metal is 3.3×1014Hz. If light of frequency 8.2×1014Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

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39. The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} H z$ . If light of frequency $8.2 \times 10^{14} H z$ is incident on this metal, the cut-off voltage for the photoelectric emission is nearly