Question

In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If \(a_0\) is the radius of the ground state orbit, \(m\) is the mass and \(e\) is charge on the electron and \(\varepsilon_0\) is the vacuum permittivity, the speed of the electron is

• A. 0
• B. \(\dfrac{e}{\sqrt{5\varepsilon_0 a_0}}\)
• C. \(\dfrac{e}{\sqrt{4\pi\varepsilon_0 a_0 m}}\)
• D. \(\dfrac{e}{\sqrt{4\pi\varepsilon_0 a_0}}\)

Hint:

In Bohr orbit, Coulomb force provides centripetal force. Equate \(\dfrac{mv^2}{r}=\dfrac{e^2}{4\pi\varepsilon_0 r^2}\) and solve for \(v\).

Answer 1

The Correct Option is C

Step 1: Write centripetal force equation.
Centripetal force required:
\[ \frac{mv^2}{r} \] Coulomb force:
\[ \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \] Step 2: Equate centripetal and coulomb force.
\[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \] Step 3: Substitute \(r=a_0\).
\[ mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{a_0} \] Step 4: Solve for speed \(v\).
\[ v^2 = \frac{e^2}{4\pi\varepsilon_0 a_0 m} \Rightarrow v = \frac{e}{\sqrt{4\pi\varepsilon_0 a_0 m}} \] Final Answer: \[ \boxed{\dfrac{e}{\sqrt{4\pi\varepsilon_0 a_0 m}}} \]