Q17 / 150 · MHT CET 2026 (17 April Shift 2) · Chemistry
Find the energy of third stationary orbit of Bohr's model of hydrogen atom if the energy of ground state is $-E$ J
- A$-\dfrac{E}{2}$ J
- B$-\dfrac{E}{4}$ J
- Correct answerC$-\dfrac{E}{9}$ J
- D$-\dfrac{E}{16}$ J
Solution
The energy of the $n$-th Bohr orbit of a hydrogen atom is given by $E_n = \dfrac{E_1}{n^2}$
Given that the energy of the ground state ($n=1$) is $E_1 = -E$
For the third stationary orbit, $n=3$
Substituting the values, we get $E_3 = \dfrac{-E}{3^2} = -\dfrac{E}{9}$ J
Answer: $-\dfrac{E}{9}$ J
Given that the energy of the ground state ($n=1$) is $E_1 = -E$
For the third stationary orbit, $n=3$
Substituting the values, we get $E_3 = \dfrac{-E}{3^2} = -\dfrac{E}{9}$ J
Answer: $-\dfrac{E}{9}$ J
Hence, the correct answer is option: 'C'.