JEE Physics 2025 Question with Solution

Given: $\frac{P_1}{P_2} = 2$, $N_1 = 2 \times 10^{15}$ photons per second, $\lambda_1 = 600 \, \text{nm}$, and $\lambda_2 = 300 \, \text{nm}$.

Find: The number of photons per second emitted by $S_2$.

The energy of one photon is

So, power can be written as

Using the ratio of powers,

The factor $hc$ cancels, so

Substitute the given wavelengths:

Now use $N_1 = 2 \times 10^{15}$:

Therefore, the number of photons per second emitted by $S_2$ is $5 \times 10^{14}$. Hence, the required numerical answer is $5$.

Given: $\frac{P_1}{P_2} = 2$, $\lambda_1 = 600 \, \text{nm}$, $\lambda_2 = 300 \, \text{nm}$, and $N_1 = 2 \times 10^{15}$.

Find: The value multiplying $10^{14}$ for source $S_2$.

Since photon energy is inversely proportional to wavelength,

So, for half the wavelength, the energy per photon doubles. Thus source $S_2$ has photons of twice the energy of source $S_1$.

Also,

Given $P_1 = 2P_2$ and $E_2 = 2E_1$,

Hence,

so

Therefore, the required answer is $5$.