As one considers orbits with higher value of n in a hydrogen atom, the electron potential energy of the atom

Options

A. decreases
B. increases
C. remain the same
D. does not increases

Answer

To solve the question regarding the behavior of electron potential energy in a hydrogen atom as the principal quantum number \( n \) increases, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Potential Energy in Hydrogen Atom:**
The potential energy (\( PE \)) of an electron in a hydrogen-like atom is given by the formula:
\[ PE = -\frac{Z e^2}{4 \pi \epsilon_0 r} \]
where \( Z \) is the atomic number, \( e \) is the charge of the electron, \( \epsilon_0 \) is the permittivity of free space, and \( r \) is the radius of the orbit.

2. **Relating Potential Energy to Total Energy:**
In a hydrogen atom, the total energy (\( E \)) is related to the potential energy by the equation:
\[ PE = 2E \]
The total energy for a hydrogen atom can be expressed as:
\[ E = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
For hydrogen, \( Z = 1 \), so:
\[ E = -\frac{13.6}{n^2} \text{ eV} \]

3. **Calculating Potential Energy:**
Substituting the expression for total energy into the potential energy equation:
\[ PE = 2E = 2 \left(-\frac{13.6}{n^2}\right) = -\frac{27.2}{n^2} \text{ eV} \]

4. **Analyzing the Effect of Increasing \( n \):**
As \( n \) increases, the term \( n^2 \) in the denominator increases, which means that the absolute value of the potential energy (which is negative) becomes smaller. Thus, the potential energy becomes less negative:
\[ PE \text{ increases (less negative)} \text{ as } n \text{ increases.} \]

5. **Conclusion:**
Therefore, as \( n \) increases, the electron potential energy of the atom increases (becomes less negative).

### Final Answer:
The electron potential energy of the atom will **increase** as one considers orbits with higher values of \( n \).