A beam of light having wavelength distributed uniformly between `450 nmto 550 nm` passes through a sample of hydrogen gas which wavelength will have the least intensity in the transition beam?

Explanation

To solve the problem, we need to determine which wavelength of light will have the least intensity after passing through a sample of hydrogen gas. The light beam has a wavelength range of 450 nm to 550 nm, and we will analyze the energy transitions in hydrogen to find the wavelength that corresponds to the least intensity.

### Step-by-Step Solution:

1. **Identify the Wavelength Range**:
The given wavelength range of the light beam is from 450 nm to 550 nm.

2. **Calculate the Energy for Minimum Wavelength (450 nm)**:
The energy associated with a wavelength can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \) eV·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) = 450 nm = \( 450 \times 10^{-9} \) m

Substituting the values:
\[ E_1 = \frac{1242}{450} \approx 2.76 \text{ eV} \]

3. **Calculate the Energy for Maximum Wavelength (550 nm)**:
Using the same formula for the maximum wavelength:
\[ E_2 = \frac{1242}{550} \approx 2.25 \text{ eV} \]

4. **Determine the Energy Range**:
The energy range of the light beam is from \( E_2 \) to \( E_1 \):
\[ \Delta E = E_1 - E_2 = 2.76 \text{ eV} - 2.25 \text{ eV} = 0.51 \text{ eV} \]

5. **Identify the Transitions in Hydrogen**:
For hydrogen, the energy levels are given by:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
The transitions that can occur are from higher energy levels (n = 3, 4, 5, ...) to n = 2 (the first excited state).

6. **Calculate Energy Differences for Transitions**:
- Transition from \( n=3 \) to \( n=2 \):
\[ E_{3 \to 2} = E_2 - E_3 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \approx 1.89 \text{ eV} \]
- Transition from \( n=4 \) to \( n=2 \):
\[ E_{4 \to 2} = E_2 - E_4 = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) \approx 2.55 \text{ eV} \]
- Transition from \( n=5 \) to \( n=2 \):
\[ E_{5 \to 2} = E_2 - E_5 = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{25} \right) \approx 2.85 \text{ eV} \]

7. **Identify the Wavelength Corresponding to the Least Intensity**:
The transition with the least intensity will correspond to the energy that falls within the range of 2.25 eV to 2.76 eV. The transition from \( n=4 \) to \( n=2 \) gives us:
\[ E = 2.55 \text{ eV} \]
Now, we convert this energy back to wavelength:
\[ \lambda = \frac{1242}{2.55} \approx 487 \text{ nm} \]

8. **Conclusion**:
The wavelength that will have the least intensity in the transition beam is approximately **487 nm**.