Correct Option: (C) $3.4 \text{ eV}, -6.8 \text{ eV}$
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
For a hydrogen atom, the total energy ($E_n$), kinetic energy ($K_n$), and potential energy ($U_n$) of an electron in the $n^{th}$ orbit are related by the Bohr model as follows:
- Total Energy: $E_n = -\frac{13.6}{n^2} \text{ eV}$
- Kinetic Energy: $K_n = -E_n = \frac{13.6}{n^2} \text{ eV}$
- Potential Energy: $U_n = 2E_n = -\frac{27.2}{n^2} \text{ eV}$
Given: The first excited state corresponds to $n = 2$.
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
Substitute $n = 2$ into the energy relations:
$$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$$Now, calculate Kinetic Energy ($K_2$):
$$K_2 = -E_2 = -(-3.4) = 3.4 \text{ eV}$$Now, calculate Potential Energy ($U_2$):
$$U_2 = 2 \times E_2 = 2 \times (-3.4) = -6.8 \text{ eV}$$Step 3 (Final Concluded Answer): [0.25 Mark]
The kinetic energy is $3.4 \text{ eV}$ and the potential energy is $-6.8 \text{ eV}$. Thus, option (C) is the correct choice.
💡 EL10 Concept Insight: Think of the electron like a satellite orbiting a planet; it needs positive kinetic energy to keep moving, but it is "trapped" by the nucleus's pull, which is why its potential energy is always negative.