JEE Physics 2024 Question with Solution

Given: Two light sources each emit power $200 \, \text{W}$. Their wavelengths are $\lambda_1 = 300 \, \text{nm}$ and $\lambda_2 = 500 \, \text{nm}$.

Find: The ratio of the number of photons emitted, $\frac{N_1}{N_2}$.

The energy of one photon is

$$E = \frac{hc}{\lambda}$$

So the power of a source is related to the number of photons emitted per second by

$$P = N \times E = N \times \frac{hc}{\lambda}$$

For the first source,

$$200 = N_1 \times \frac{hc}{300 \times 10^{-9}}$$

Hence,

$$N_1 = \frac{200 \cdot 300 \times 10^{-9}}{hc}$$

For the second source,

$$200 = N_2 \times \frac{hc}{500 \times 10^{-9}}$$

Hence,

$$N_2 = \frac{200 \cdot 500 \times 10^{-9}}{hc}$$

Now take the ratio:

$$\frac{N_1}{N_2} = \frac{\frac{200 \cdot 300 \times 10^{-9}}{hc}}{\frac{200 \cdot 500 \times 10^{-9}}{hc}} = \frac{300}{500} = \frac{3}{5}$$

Therefore, the ratio of the number of photons emitted is $3:5$. The correct option is D.

Given: Both sources have the same power $P = 200 \, \text{W}$.

Find: The ratio $N_1:N_2$ for wavelengths $300 \, \text{nm}$ and $500 \, \text{nm}$.

Since photon energy is

$$E = \frac{hc}{\lambda}$$

and power is

$$P = N E$$

for fixed power, the number of photons is proportional to wavelength:

$$N \propto \lambda$$

Therefore,

$$\frac{N_1}{N_2} = \frac{\lambda_1}{\lambda_2} = \frac{300}{500} = \frac{3}{5}$$

So the required ratio is $3:5$, and the correct option is D.