Question

The ratio of the kinetic energies of the electrons in the third and fourth excited states of hydrogen atom is

• A. 4:3
• B. 16:9
• C. 25:16
• D. 5:4

Hint:

Remember: $n$-th excited state means the principal quantum number is $(n+1)$. KE $\propto 1/n^2$, PE $\propto -1/n^2$, TE $\propto -1/n^2$.

Answer 1

The Correct Option is C

Step 1: Identify Quantum Numbers: Ground state: $n=1$. Third excited state: $n_1 = 3 + 1 = 4$. Fourth excited state: $n_2 = 4 + 1 = 5$.
Step 2: Kinetic Energy Relation: For a hydrogen atom, the kinetic energy $K_n$ in the $n$-th orbit is inversely proportional to $n^2$. $K_n \propto \frac{1}{n^2}$.
Step 3: Calculation Process Ratio: \[ \frac{K_{3rd\_exc}}{K_{4th\_exc}} = \frac{K_4}{K_5} = \frac{1/4^2}{1/5^2} = \frac{5^2}{4^2} = \frac{25}{16} \]