Two sources A and B have same power . The wavelength of radiation of A is `lambda_a` and that of B is `lambda_b`.The number of photons emitted per second by A and B are `n_a` and `n_b` respectively, then

Options

A. `lambda_a gt lambda_b`
B. if `lambda_a gt lambda_b , n_a lt n_b`
C. If `lambda_a lt lambda_b, n_a lt n_b`
D. if `lambda_a lt lambda_b , n_a =n_b`

Answer

To solve the problem, we need to analyze the relationship between the power of the sources, the wavelength of the emitted radiation, and the number of photons emitted per second. Let's break it down step by step:

### Step 1: Understand the relationship between power, energy, and number of photons
The power (P) of a source is defined as the energy emitted per unit time. For a source emitting photons, the power can be expressed as:
\[ P = n \cdot E \]
where:
- \( n \) is the number of photons emitted per second,
- \( E \) is the energy of one photon.

### Step 2: Express the energy of a photon in terms of wavelength
The energy of a photon can be expressed using its wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light.

### Step 3: Write the power equations for both sources
For source A:
\[ P_A = n_A \cdot E_A = n_A \cdot \frac{hc}{\lambda_A} \]
For source B:
\[ P_B = n_B \cdot E_B = n_B \cdot \frac{hc}{\lambda_B} \]

Since both sources have the same power:
\[ P_A = P_B \]

### Step 4: Set the power equations equal to each other
From the above equations, we can write:
\[ n_A \cdot \frac{hc}{\lambda_A} = n_B \cdot \frac{hc}{\lambda_B} \]

### Step 5: Cancel out common terms
Since \( hc \) is common in both sides, we can cancel it out:
\[ n_A \cdot \frac{1}{\lambda_A} = n_B \cdot \frac{1}{\lambda_B} \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ \frac{n_A}{n_B} = \frac{\lambda_A}{\lambda_B} \]

### Step 7: Analyze the relationship
From the equation \( \frac{n_A}{n_B} = \frac{\lambda_A}{\lambda_B} \), we can conclude:
- If \( \lambda_A > \lambda_B \), then \( n_A < n_B \) (source A emits fewer photons than source B).
- If \( \lambda_A < \lambda_B \), then \( n_A > n_B \) (source A emits more photons than source B).

### Conclusion
Thus, the number of photons emitted per second is inversely proportional to the wavelength of the emitted radiation. If the wavelength of A is greater than that of B, then the number of photons emitted by A is less than that emitted by B.

### Final Answer
If \( \lambda_A > \lambda_B \), then \( n_A < n_B \).