Q137 / 180 · KCET 2018 · Physics
The number of photons falling per second on a completely darkened plate to produce a force
of \( 6.62 \times 10^{-5} \mathrm{~N} \) is ' \( n \) '. If the wavelength of the light falling is \( 5 \times 10^{-7} \mathrm{~m} \), then \( n= \)
\( \times 10^{22} \).
\( \left(h=6.62 \times 10^{-34}\right) \mathrm{J}_{\mathrm{S}} \)
of \( 6.62 \times 10^{-5} \mathrm{~N} \) is ' \( n \) '. If the wavelength of the light falling is \( 5 \times 10^{-7} \mathrm{~m} \), then \( n= \)
\( \times 10^{22} \).
\( \left(h=6.62 \times 10^{-34}\right) \mathrm{J}_{\mathrm{S}} \)
- A\( 11 \)
- Correct answerB\( 05 \)
- C\( 0.2 \)
- D\( 3.3 \)
Solution
Given, force, $F=6.62 \times 10^{-5} \mathrm{~N} ;$ wavelength,$\lambda=5 \times 10^{-7} \mathrm{~m}$
Now, $C \times F=\frac{h c}{\lambda} \times \frac{n}{t}$
or $F=\frac{h n}{\lambda t}$
$\Rightarrow \frac{n}{t}=\frac{F \lambda}{h}=\frac{6.62 \times 10^{-5}}{6.62 \times 10^{-34}} \times 5 \times 10^{-7}=5 \times 10^{22}$
Therefore, number of photons falling per second $=5 \times 10^{22}$
Hence, the correct answer is option: 'B'.