Q148 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
In Bohr's atomic model, the energy of the electron is 'E' in the second orbit of hydrogen atom. So the energy of the electron in the third orbit of helium atom $(Z = 2)$ will be
- A$\dfrac{16\,E}{3}$
- Correct answerB$\dfrac{16\,E}{9}$
- C$\dfrac{4\,E}{3}$
- D$\dfrac{4\,E}{9}$
Solution
The energy of an electron in the $n$-th orbit of a hydrogen-like atom is given by $E_n \propto \dfrac{Z^2}{n^2}$.
For the second orbit of a hydrogen atom ($Z=1, n=2$):
$E = k \dfrac{1^2}{2^2} = \dfrac{k}{4} \Rightarrow k = 4E$
For the third orbit of a helium atom ($Z=2, n=3$):
$E' = k \dfrac{2^2}{3^2} = k \dfrac{4}{9}$
Substituting $k = 4E$:
$E' = 4E \times \dfrac{4}{9} = \dfrac{16E}{9}$
Answer: $\dfrac{16\,E}{9}$
For the second orbit of a hydrogen atom ($Z=1, n=2$):
$E = k \dfrac{1^2}{2^2} = \dfrac{k}{4} \Rightarrow k = 4E$
For the third orbit of a helium atom ($Z=2, n=3$):
$E' = k \dfrac{2^2}{3^2} = k \dfrac{4}{9}$
Substituting $k = 4E$:
$E' = 4E \times \dfrac{4}{9} = \dfrac{16E}{9}$
Answer: $\dfrac{16\,E}{9}$
Hence, the correct answer is option: 'B'.