Question

According to Bohr’s postulate, the centripetal force \(F\) necessary for the electron of mass \(m\) in a hydrogen atom to revolve in the \(n^{\text{th}}\) circular orbit round the nucleus as the centre is given by \([e =\) charge on electron, \(h =\) Planck’s constant, \(\varepsilon_0 =\) permittivity of free space\(]\)

• A. \(\dfrac{\pi m^{2} e^{6}}{4 \varepsilon_{0}^{3} h^{4} n^{2}}\)
• B. \(\dfrac{\pi m^{2} e^{6}}{4 \varepsilon_{0}^{3} h^{4} n^{4}}\)
• C. \(\dfrac{\pi m^{2} e^{4}}{4 \varepsilon_{0}^{2} h^{4} n^{4}}\)
• D. \(\dfrac{\pi m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)

Hint:

In Bohr’s model, centripetal force is provided by electrostatic attraction between nucleus and electron.

Answer 1

The Correct Option is B

Step 1: Expression for centripetal force.
For circular motion of electron:
\[ F = \frac{mv^{2}}{r} \]
Step 2: Use Bohr’s postulates.
Radius of \(n^{\text{th}}\) orbit:
\[ r_n = \frac{\varepsilon_0 h^{2} n^{2}}{\pi m e^{2}} \] Velocity of electron in \(n^{\text{th}}\) orbit:
\[ v_n = \frac{e^{2}}{2 \varepsilon_0 h n} \]
Step 3: Substitute \(v_n\) and \(r_n\) in force equation.
\[ F = \frac{m \left(\dfrac{e^{2}}{2 \varepsilon_0 h n}\right)^{2}}{\dfrac{\varepsilon_0 h^{2} n^{2}}{\pi m e^{2}}} \]
Step 4: Simplify.
\[ F = \frac{\pi m^{2} e^{6}}{4 \varepsilon_0^{3} h^{4} n^{4}} \]
Step 5: Conclusion.
The required centripetal force is \(\dfrac{\pi m^{2} e^{6}}{4 \varepsilon_{0}^{3} h^{4} n^{4}}\).