JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Modern Physics

The ratio of the power of a light source

S_{1}

to that the light source

S_{2}

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}

Answer: 5

Solution

The power radiated by a source equals the energy carried by each photon times the number of photons emitted per second:

P = n \cdot \frac{h c}{λ} .

For the two sources,

\frac{P _{1}}{P _{2}} = \frac{n _{1} ( h c / λ _{1} )}{n _{2} ( h c / λ _{2} )} = \frac{n _{1}}{n _{2}} \cdot \frac{λ _{2}}{λ _{1}} .

Given

P_{1} / P_{2} = 2

n_{1} = 2 \times 1 0^{15}

λ_{1} = 600 nm

λ_{2} = 300 nm

2 = \frac{2 \times 1 0 ^{15}}{n _{2}} \times \frac{300}{600} = \frac{2 \times 1 0 ^{15}}{n _{2}} \times \frac{1}{2} = \frac{1 0 ^{15}}{n _{2}} .

n_{2} = \frac{1 0 ^{15}}{2} = 5 \times 1 0^{14} .

So the number of photons per second emitted by

S_{2}

5 \times 1 0^{14}

, i.e. the required integer is

5

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer