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: Use angular momentum:
$mvr = \dfrac{nh}{2\pi}$, so $v = \dfrac{nh}{2\pi mr}$.

Step 2: Period:
$T = \dfrac{2\pi r}{v} = \dfrac{4\pi^2mr^2}{nh}$. Put $r = \dfrac{\varepsilon_0n^2h^2}{\pi me^2}$: $r^2 = \dfrac{\varepsilon_0^2n^4h^4}{\pi^2m^2e^4}$.

Step 3: Combine:
$T = \dfrac{4\pi^2m}{nh}\cdot\dfrac{\varepsilon_0^2n^4h^4}{\pi^2m^2e^4} = \dfrac{4\varepsilon_0^2n^3h^3}{me^4}$, option (A).

Final Answer:
The orbital period is 4 eps0^2 n^3 h^3 / (m e^4). \[ \boxed{\text{(A) }\dfrac{4\varepsilon_0^2n^3h^3}{me^4}} \]