JEE Main 2025 Jan 24 Shift 2, Physics Q23: Introduction and Photoelectric Effect

JEE Main2025Jan 24, Shift 2Physics

The ratio of the power of a light source $S_{1}$ to that the light source $S_{2}$ is $2$ . $S_{1}$ is emitting $2 \times 1 0^{15}$ photons per second at 600 nm. If the wavelength of the source $S_{2}$ is 300 nm, then the number of photons per second emitted by $S_{2}$ is ______ $\times 1 0^{14}$ .

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Solution

Power from a source emitting $n$ photons per second at wavelength $λ$ :

$P = n \cdot E_{photon} = n \cdot \frac{h c}{λ}$

Taking the ratio:

$\frac{P _{1}}{P _{2}} = \frac{n _{1} / λ _{1}}{n _{2} / λ _{2}} = \frac{n _{1} λ _{2}}{n _{2} λ _{1}}$

Plug in $P_{1} / P_{2} = 2$ , $λ_{1} = 600$ nm, $λ_{2} = 300$ nm, $n_{1} = 2 \times 1 0^{15}$ :

$2 = \frac{( 2 \times 1 0 ^{15} ) \cdot 300}{n _{2} \cdot 600} = \frac{1 0 ^{15}}{n _{2}}$

$n_{2} = \frac{1 0 ^{15}}{2} = 5 \times 1 0^{14} photons/s$

Answer: 5

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JEE Main 2025 Jan 24 Shift 2, Physics Q23: Introduction and Photoelectric Effect

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