Q102 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
In Bohr's theory of hydrogen atom, 'r' is the radius of the orbit, 'V' is the speed of electron and E is the total energy of the electron. Out of the following which physical quantity is inversely proportional to principal quantum number?
- Correct answerA$\dfrac{1}{Vr}$
- B$Vr$
- C$V^2r$
- D$Vr^2$
Solution
According to Bohr's theory of hydrogen atom:
Radius of the $n$-th orbit, $r \propto n^2$
Speed of the electron in the $n$-th orbit, $V \propto \dfrac{1}{n}$
Now, evaluating the given options:
For $Vr$, we have $Vr \propto \left(\dfrac{1}{n}\right)(n^2) \Rightarrow Vr \propto n$
For $\dfrac{1}{Vr}$, we have $\dfrac{1}{Vr} \propto \dfrac{1}{n}$
For $V^2r$, we have $V^2r \propto \left(\dfrac{1}{n^2}\right)(n^2) \Rightarrow V^2r \propto n^0$
For $Vr^2$, we have $Vr^2 \propto \left(\dfrac{1}{n}\right)(n^4) \Rightarrow Vr^2 \propto n^3$
Thus, the quantity $\dfrac{1}{Vr}$ is inversely proportional to the principal quantum number $n$.
Answer: $\dfrac{1}{Vr}$
Radius of the $n$-th orbit, $r \propto n^2$
Speed of the electron in the $n$-th orbit, $V \propto \dfrac{1}{n}$
Now, evaluating the given options:
For $Vr$, we have $Vr \propto \left(\dfrac{1}{n}\right)(n^2) \Rightarrow Vr \propto n$
For $\dfrac{1}{Vr}$, we have $\dfrac{1}{Vr} \propto \dfrac{1}{n}$
For $V^2r$, we have $V^2r \propto \left(\dfrac{1}{n^2}\right)(n^2) \Rightarrow V^2r \propto n^0$
For $Vr^2$, we have $Vr^2 \propto \left(\dfrac{1}{n}\right)(n^4) \Rightarrow Vr^2 \propto n^3$
Thus, the quantity $\dfrac{1}{Vr}$ is inversely proportional to the principal quantum number $n$.
Answer: $\dfrac{1}{Vr}$
Hence, the correct answer is option: 'A'.