An electron in a hydrogen atom is excited from the ground state to the n = 4 state. Predict which of the following statements are true or false :
(i) `n = 4` is the first excited state.
(ii) It takes more energy to ionize (remove) the electron from n = 14 than in the ground state.
(iii) The wavelength of light emitted when the electron drops from `n = 4` to `n = 2` is longer than that from `n = 4` to `n = 1`.
(iv) The wavelength which the atom absorbs in going from `n = 1` to `n = 4` is the same as emitted when it goes from `n = 4` to `n = 1`.
(v) The electron is farther from the nucleus (on average) in n = 4 than in the ground state.

Answer

To determine the truth value of each statement regarding the hydrogen atom when an electron is excited from the ground state to the n = 4 state, we will analyze each statement one by one.

### Solution Steps:

1. **Statement (i): `n = 4` is the first excited state.**
- The ground state of hydrogen corresponds to n = 1. The first excited state is n = 2, the second excited state is n = 3, and so on. Therefore, n = 4 is actually the third excited state.
- **Conclusion**: **False**

2. **Statement (ii): It takes more energy to ionize (remove) the electron from n = 14 than in the ground state.**
- The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
For n = 1 (ground state), the energy is -13.6 eV. For n = 14, the energy is:
\[ E_{14} = -\frac{13.6}{14^2} \approx -0.0693 \text{ eV} \]
Ionization energy from n = 1 is 13.6 eV, while from n = 14 it is approximately 0.0693 eV. Thus, it takes less energy to ionize from n = 14 than from the ground state.
- **Conclusion**: **False**

3. **Statement (iii): The wavelength of light emitted when the electron drops from `n = 4` to `n = 2` is longer than that from `n = 4` to `n = 1`.**
- The energy difference for transitions is given by:
\[ \Delta E = E_{n_1} - E_{n_2} \]
For n = 4 to n = 2:
\[ \Delta E_{4 \to 2} = E_2 - E_4 \]
For n = 4 to n = 1:
\[ \Delta E_{4 \to 1} = E_1 - E_4 \]
Since the transition from n = 4 to n = 1 involves a larger energy difference than from n = 4 to n = 2, the wavelength emitted (which is inversely proportional to energy) will be shorter for n = 4 to n = 1.
- **Conclusion**: **False**

4. **Statement (iv): The wavelength which the atom absorbs in going from `n = 1` to `n = 4` is the same as emitted when it goes from `n = 4` to `n = 1`.**
- The energy absorbed when moving from n = 1 to n = 4 is equal to the energy emitted when moving from n = 4 to n = 1. Thus, the wavelengths will be the same because they correspond to the same energy change.
- **Conclusion**: **True**

5. **Statement (v): The electron is farther from the nucleus (on average) in n = 4 than in the ground state.**
- The radius of the nth orbit in a hydrogen atom is given by:
\[ r_n = 0.529 \, \frac{n^2}{Z} \text{ angstroms} \]
Since the radius is proportional to \( n^2 \), the electron in n = 4 will be farther from the nucleus than in n = 1.
- **Conclusion**: **True**

### Final Answers:
- (i) False
- (ii) False
- (iii) False
- (iv) True
- (v) True