MHT CET · Physics · Dual Nature of Matter
When light of wavelength ' \(\lambda\) ' is incident on photosensitive surface, photons of power
'P' are emitted. The number of photons (n) emitted in 't' second is
(h= Planck's constant, \(\mathrm{c}=\) velocity of light in vacuum)
- A \(\frac{h \mathrm{C}}{\mathrm{P} \lambda t}\)
- B \(\frac{P \lambda t}{h_{C}}\)
- C \(\frac{\mathrm{P} \lambda}{\operatorname{htc}}\)
- D \(\frac{h \mathrm{P}}{\lambda t \mathbf{C}}\)
Answer & Solution
Correct Answer
(B) \(\frac{P \lambda t}{h_{C}}\)
Step-by-step Solution
Detailed explanation
Energy of each photon \(=\frac{h c}{\lambda}\)
If \(n\) photons are emitted in time \(t\) then the energy emitted in time \(t\) is \(\frac{\text { nhc }}{\lambda}\)
\(\begin{aligned} \text { Power P } &=\text { Energy emitted per second } \\ &=\frac{\mathrm{nhc}}{\lambda \mathrm{t}} \\ \therefore \mathrm{n} &=\frac{\mathrm{P} \lambda \mathrm{t}}{\mathrm{hc}} \end{aligned}\)
If \(n\) photons are emitted in time \(t\) then the energy emitted in time \(t\) is \(\frac{\text { nhc }}{\lambda}\)
\(\begin{aligned} \text { Power P } &=\text { Energy emitted per second } \\ &=\frac{\mathrm{nhc}}{\lambda \mathrm{t}} \\ \therefore \mathrm{n} &=\frac{\mathrm{P} \lambda \mathrm{t}}{\mathrm{hc}} \end{aligned}\)
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