The ground state energy of hydrogen atom is -13.6 eV (i) What are the potential energy and K.E. of electron is 3rd excited state?
(ii) If the electron jumped to the ground state form the third excited state, calculate the frequency of photon emitted.

Explanation

To solve the problem step by step, we will break it down into two parts as stated in the question.

### Part (i): Potential Energy and K.E. of Electron in 3rd Excited State

1. **Identify the Energy Formula**:
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

2. **Determine n for the 3rd Excited State**:
The 3rd excited state corresponds to \( n = 4 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), the second excited state is \( n = 3 \), and the third excited state is \( n = 4 \)).

3. **Calculate the Energy for n = 4**:
Substitute \( n = 4 \) into the energy formula:
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

4. **Calculate the Kinetic Energy (K.E.)**:
The kinetic energy of the electron is equal to the negative of the total energy:
\[ K.E. = -E_4 = -(-0.85 \, \text{eV}) = 0.85 \, \text{eV} \]

5. **Calculate the Potential Energy (P.E.)**:
The potential energy is twice the kinetic energy but negative:
\[ P.E. = -2 \times K.E. = -2 \times 0.85 \, \text{eV} = -1.70 \, \text{eV} \]

### Summary of Part (i):
- K.E. = 0.85 eV
- P.E. = -1.70 eV

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### Part (ii): Frequency of Photon Emitted when Jumping to Ground State

1. **Determine the Energy of the Ground State**:
The energy of the ground state (n = 1) is:
\[ E_1 = -13.6 \, \text{eV} \]

2. **Calculate the Change in Energy (ΔE)**:
The change in energy when the electron jumps from the 3rd excited state (n = 4) to the ground state (n = 1) is:
\[ \Delta E = E_1 - E_4 = -13.6 \, \text{eV} - (-0.85 \, \text{eV}) = -13.6 + 0.85 = -12.75 \, \text{eV} \]

3. **Convert ΔE to Joules**:
To find the frequency, we need to convert energy from eV to Joules. Using \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 12.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.04 \times 10^{-18} \, \text{J} \]

4. **Use Planck's Equation to Find Frequency (ν)**:
The relationship between energy and frequency is given by:
\[ \Delta E = h \nu \]
where \( h = 6.63 \times 10^{-34} \, \text{J s} \). Rearranging gives:
\[ \nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \approx 3.08 \times 10^{15} \, \text{Hz} \]

### Summary of Part (ii):
- Frequency of photon emitted, \( \nu \approx 3.08 \times 10^{15} \, \text{Hz} \)

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