The ground state energy of hydrogen atom is `-13.6eV`. If the electron jumps to the ground state from the `3^("rd")` excited state, the wavelength of the emitted photon is

Options

A. `875Å`
B. `1052Å`
C. `752Å`
D. `1026Å`

Answer

To find the wavelength of the emitted photon when an electron in a hydrogen atom jumps from the third excited state to the ground state, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Energy Levels:**
- The ground state of hydrogen corresponds to \( n = 1 \).
- The third excited state corresponds to \( n = 4 \) (since the first excited state is \( n = 2 \), the second is \( n = 3 \), and the third is \( n = 4 \)).

2. **Calculate the Energy of Each State:**
- The energy of an electron in a hydrogen atom at level \( n \) is given by the formula:
\[ E_n = -\frac{E_0}{n^2} \]
where \( E_0 = 13.6 \, \text{eV} \).
- For the ground state (\( n = 1 \)):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
- For the third excited state (\( n = 4 \)):
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \]

3. **Calculate the Energy Released During the Transition:**
- The energy released when the electron transitions from \( n = 4 \) to \( n = 1 \) is:
\[ \Delta E = E_1 - E_4 = (-13.6) - (-0.85) = -13.6 + 0.85 = -12.75 \, \text{eV} \]
- Thus, the energy released is \( 12.75 \, \text{eV} \).

4. **Convert Energy to Joules:**
- To convert electron volts to joules, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 12.75 \times 1.6 \times 10^{-19} \, \text{J} = 2.04 \times 10^{-18} \, \text{J} \]

5. **Use Planck's Equation to Find Wavelength:**
- Planck's equation relates energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
- Rearranging for wavelength \( \lambda \):
\[ \lambda = \frac{hc}{E} \]
- Where \( h = 6.626 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2.04 \times 10^{-18}} \]

6. **Calculate the Wavelength:**
- Performing the calculation:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{2.04 \times 10^{-18}} \approx 9.74 \times 10^{-8} \, \text{m} \]
- Converting to nanometers (1 nm = \( 10^{-9} \, \text{m} \)):
\[ \lambda \approx 97.4 \, \text{nm} \]

### Final Answer:
The wavelength of the emitted photon is approximately \( 97.4 \, \text{nm} \).