Question

Using Bohr's atomic model, the orbital period of electron in hydrogen atom in the \(n^{\text{th}}\) orbit is (\(ε_0\)=permittivity of free space, h=Planck's constant, m = mass of electron, e = electronic charge)

• A. \(\frac{4\,ε_0^2n^3h^3}{me^4}\)
• B. \(\frac{2ε_0^2n^2h^3}{me^2}\)
• C. \(\frac{4ε_0^2n^2h^3}{me^2}\)
• D. \(\frac{2ε_0n^3h^3}{me^4}\)

Hint:

Period equals orbit circumference over speed; use Bohr radius and speed formulas.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Period \(T = \dfrac{2\pi r_n}{v_n}\). From Bohr's theory:
\(r_n = \dfrac{\varepsilon_0n^2h^2}{\pi me^2}\) and \(v_n = \dfrac{e^2}{2\varepsilon_0nh}\).

Step 2: Substitute:
\[ T = 2\pi\cdot\frac{\varepsilon_0n^2h^2}{\pi me^2}\cdot\frac{2\varepsilon_0nh}{e^2} \]

Step 3: Simplify:
The factor \(\pi\) cancels:
\[ T = \frac{4\varepsilon_0^2n^3h^3}{me^4} \]

Step 4: Check:
Option (A). The power of \(n\) is 3, as expected since \(T\propto n^3\) (radius \(n^2\) over speed \(1/n\)). Options (B) and (C) have \(n^2\) and \(e^2\), which do not fit.

Final Answer:
T = 2 pi r / v with the Bohr radius and speed. \[ \boxed{\text{(A) }\dfrac{4\varepsilon_0^2n^3h^3}{me^4}} \]