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 \(]\) :
- A \(\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}\)
- B \(\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}\)
- C \(\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}\)
- D \(\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}\)
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
\(\begin{array}{l}
r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\
\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}
\end{array}\)
\(\begin{array}{l}
m v r=\frac{n h}{2 \pi} \quad \text { So, } v=\frac{n h}{2 \pi m r} \\
\mathrm{KE}=\frac{1}{2} m v^{2} \quad \text { So, } \mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2}
\end{array}\)
\(\text { Since, } r=\frac{a_{\mathrm{o}} \times n^{2}}{z}\)
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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