Q123 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it
- Awill decrease
- Correct answerBwill increase
- Cwill remain same
- Dwill be zero
Solution
According to Bohr's model, the velocity of an electron in the $n$-th orbit is given by $v \propto \dfrac{1}{n}$.
The de-Broglie wavelength of the electron is $\lambda = \dfrac{h}{mv}$.
Substituting the proportionality of velocity, we get $\lambda \propto n$.
The ground state corresponds to $n = 1$ and the third excited state corresponds to $n = 4$.
Since $n$ increases from $1$ to $4$, the de-Broglie wavelength of the electron will increase.
Answer: will increase
The de-Broglie wavelength of the electron is $\lambda = \dfrac{h}{mv}$.
Substituting the proportionality of velocity, we get $\lambda \propto n$.
The ground state corresponds to $n = 1$ and the third excited state corresponds to $n = 4$.
Since $n$ increases from $1$ to $4$, the de-Broglie wavelength of the electron will increase.
Answer: will increase
Hence, the correct answer is option: 'B'.