1. Energy of one photon

For light of wavelength $\lambda$, the energy of one photon is

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

So, photon energy is inversely proportional to wavelength.

2. Number of photons emitted per second

If a source emits power $P$, then the number of photons emitted per second is

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

Thus,

$N \propto \lambda$

for the same power $P$.

3. Apply to the two wavelengths

Let $N_1$ be the number of photons for $\lambda_1 = 300\,\text{nm}$, and $N_2$ be the number of photons for $\lambda_2 = 500\,\text{nm}$.

Since both sources have the same power,

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

So the required ratio is

$N_{300\,\text{nm}} : N_{500\,\text{nm}} = 3:5$

4. Check options

• A: $5:3$ ❌
• B: $3:5$ ✅
• C: $1:5$ ❌
• D: $1:3$ ❌

Therefore, the correct answer is B.