Question

An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?

• A. Kinetic energy, potential energy and total energy decreases
• B. Kinetic energy and total energy decreased but potential energy increases
• C. Kinetic energy increases but potential energy and total energy decreases
• D. Kinetic energy decreases, potential energy increases but total energy remains the same.

Hint:

Remember the classic Bohr orbit relationships: $T.E. = -K.E.$ and $P.E. = 2 \times T.E.$. As an electron falls closer to the nucleus, it loses total energy (it becomes more negative/stable) and loses potential energy, while speeding up to maintain its orbit, thereby increasing its kinetic energy.

Answer 1

The Correct Option is C

Step 1: Use the virial theorem shortcut.
Rather than recall three separate formulas, remember the elegant energy relations for any bound electron in a Coulomb field: kinetic energy is positive, and total energy is negative and equal in size to the kinetic energy.
Step 2: State the ratios.
For a circular orbit, $K = -E$ and $U = 2E$, where $E$ is the (negative) total energy. Numerically, if $E = -13.6\,\text{eV}$ for the ground state, then $K = +13.6\,\text{eV}$ and $U = -27.2\,\text{eV}$.
Step 3: Track the transition.
Going from an excited state to the ground state means $n$ falls, so $E$ becomes more negative (energy is radiated as a photon).
Step 4: Effect on total energy.
Since $E$ becomes more negative, the total energy decreases.
Step 5: Effect on potential energy.
Because $U = 2E$, $U$ also becomes more negative, so the potential energy decreases.
Step 6: Effect on kinetic energy and conclude.
Because $K = -E$ and $E$ drops, $K$ rises. So kinetic energy increases while potential and total energy decrease. \[ \boxed{K\uparrow,\ U\downarrow,\ E\downarrow} \]