Question

The total energy of an electron in the first excited state of hydrogen is about -3.4eV. Its kinetic energy in this state is:

• A. - 3.4 eV
• B. - 1.7 eV
• C. 1.7 eV
• D. 3.4 eV

Answer 1

The Correct Option is D

 The question asks us to determine the kinetic energy of an electron in the first excited state of hydrogen, given that its total energy is -3.4 eV. In atomic physics, particularly for the hydrogen atom, the total energy of an electron in any given energy state (n) is given by the formula:

\(E_n = -\frac{13.6}{n^2} \text{ eV}\)

This formula represents the potential energy, which is inherently negative, reflecting the bounded state of the electron in the atom. The kinetic energy, on the other hand, is a positive value and is numerically equal to the absolute value of the total energy in that state. This is derived as follows:

1. The total energy \(E\) of the electron is the sum of its kinetic energy \(T\) and potential energy \(V\), expressed as:
   • \(E = T + V\)
2. The potential energy \(V\) is twice the total energy (and negative), i.e., \(V = 2E\).
3. Since the total energy \(E = -3.4 \text{ eV}\) for the first excited state (n=2), the kinetic energy \(T\) is:
   • \(T = -E = 3.4 \text{ eV}\)

Thus, the kinetic energy of the electron in the first excited state of hydrogen is \(3.4 \text{ eV}\).

Let us now evaluate the given options:

• -3.4 eV: Incorrect. Kinetic energy should be positive.
• -1.7 eV: Incorrect. Does not match the calculated kinetic energy value.
• 1.7 eV: Incorrect. This is incorrect as the kinetic energy equals the absolute value of the total energy.
• 3.4 eV: Correct. Matches the calculated value for kinetic energy.

Hence, the correct answer is 3.4 eV.