When electronic transition occurs from higher energy state to lower energy state with energy difference equal to `Delta E` electron volts , the wavelength of the line emitted is approxmately equal to

Options

A. `(12375)/(DeltaE)` m
B. `(12375)/(DeltaE)` nm
C. `(12375)/(DeltaE)` pm
D. `(12375)/(DeltaE)Å`

Answer

To find the wavelength of the emitted line when an electron transitions from a higher energy state to a lower energy state with an energy difference of ΔE electron volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy of a photon emitted during an electronic transition is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert energy from electron volts to joules**:
Since the energy difference ΔE is given in electron volts (eV), we need to convert it to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]
Therefore, the energy in joules is:
\[ E = \Delta E \times 1.602 \times 10^{-19} \, \text{J} \]

3. **Rearrange the energy-wavelength equation**:
To find the wavelength \(\lambda\), we can rearrange the equation:
\[ \lambda = \frac{hc}{E} \]

4. **Substitute the values**:
Substitute \(E\) with the expression we derived:
\[ \lambda = \frac{hc}{\Delta E \times 1.602 \times 10^{-19}} \]
Now substituting the values of \(h\) and \(c\):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{\Delta E \times 1.602 \times 10^{-19}} \]

5. **Calculate the numerical value**:
Calculate the numerator:
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^{8} \approx 1.9878 \times 10^{-25} \, \text{Js m} \]
Now substituting this value into the equation for \(\lambda\):
\[ \lambda \approx \frac{1.9878 \times 10^{-25}}{\Delta E \times 1.602 \times 10^{-19}} \]

6. **Simplify the expression**:
\[ \lambda \approx \frac{1.237 \times 10^{-6}}{\Delta E} \text{ meters} \]
or converting to nanometers (1 nm = \(10^{-9}\) m):
\[ \lambda \approx \frac{1239.8}{\Delta E} \text{ nm} \]

### Final Answer:
The wavelength of the emitted line is approximately:
\[ \lambda \approx \frac{1239.8}{\Delta E} \text{ nm} \]