If `Delta E` is the energy emitted in electron volts when an electronic transition occurs from higher energy level to a lower energy level in H-atom, the wavelength of the line produced is approximately equal to

Options

A. `(19800)/(DeltaE) Å`
B. `(12375)/(DeltaE) Å`
C. `(13600)/(Delta E) Å`
D. `(21800)/(Delta E) Å`

Explanation

To find the wavelength of the emitted light when an electron transitions from a higher energy level to a lower energy level in a hydrogen atom, we can use the relationship between energy and wavelength given by the equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(E\) is the energy emitted (in joules),
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength (in meters).

Since the energy \(\Delta E\) is given in electron volts (eV), we need to convert it to joules. The conversion factor is:

\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]

### Step-by-Step Solution:

1. **Convert Energy from eV to Joules**:
\[ E (\text{in J}) = \Delta E (\text{in eV}) \times 1.602 \times 10^{-19} \, \text{J/eV} \]

2. **Rearrange the Energy-Wavelength Equation**:
\[ \lambda = \frac{hc}{E} \]

3. **Substitute the Values**:
- Substitute \(h = 6.626 \times 10^{-34} \, \text{J s}\),
- Substitute \(c = 3.00 \times 10^8 \, \text{m/s}\),
- Substitute \(E\) from step 1.

4. **Calculate the Wavelength**:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s})}{\Delta E (\text{in eV}) \times 1.602 \times 10^{-19} \, \text{J/eV}} \]

5. **Final Calculation**:
- Perform the calculation to find \(\lambda\).