Question

If the electron in a hydrogen atom moves from the ground state orbit to the $5^{\text{th}}$ orbit, then the potential energy of the electron

• A. is increased
• B. is zero
• C. is decreased
• D. remains unchanged

Hint:

Think of gravity as an analogy: pulling an object further away from its attractive center always increases its potential energy! Because the value moves from a deep negative well closer toward zero, it is mathematically increasing.

Answer 1

The Correct Option is A

The electrostatic potential energy ($U$) of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by: $$U = -\frac{1}{4\pi\epsilon_0} \cdot \frac{e^2}{r_n} = \frac{-27.2}{n^2}\ \text{eV}$$ * In the ground state ($n = 1$), the potential energy is $U_1 = -27.2\ \text{eV}$. * In the $5^{\text{th}}$ orbit ($n = 5$), the potential energy is $U_5 = \frac{-27.2}{25} = -1.088\ \text{eV}$. Since $-1.088\ \text{eV}$ is a larger (less negative) value than $-27.2\ \text{eV}$, the potential energy has increased as the electron moved further away from the attractive pull of the positive nucleus.
Final Answer:
The potential energy of the electron is increased, which corresponds to option (A).