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JEE Advanced · Chemistry · 2. Atomic Structure

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is \(\left[a_{0}\right.\) is Bohr radius \(]\) :

  1. A \(\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}\)
  2. B \(\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}\)
  3. C \(\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}\)
  4. D \(\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}\)

Step-by-step Solution

Detailed explanation

As per Bohr's postulate,

\(\begin{array}{l}

m v r=\frac{n h}{2 \pi} \quad \text { So, } v=\frac{n h}{2 \pi m r} \\

\mathrm{KE}=\frac{1}{2} m v^{2} \quad \text { So, } \mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2}

\end{array}\)

\(\text { Since, } r=\frac{a_{\mathrm{o}} \times n^{2}}{z}\)

So, for \(2^{\text {nd }}\) Bohr orbit

\(\begin{array}{l}

r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\

\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}

\end{array}\)
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