Atomic Structure Result Question 17-1
17. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ $a_0$ is Bohr radius]
(2012)
(a) $\dfrac{h^2}{4 \pi^2 m a_0^2}$
(b) $\dfrac{h^2}{16 \pi^2 m a_0^2}$
(c) $\dfrac{h^2}{32 \pi^2 m a_0^2}$
(d) $\dfrac{h^2}{64 \pi^2 m a_0^2}$
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Answer:
Correct Answer: 17. ( c )
Solution:
- According to Bohr’s model,
$\begin{aligned} & m v r=\dfrac{n h}{2 \pi} \Rightarrow(m v)^2=\dfrac{n^2 h^2}{4 \pi^2 r^2} \\ \Rightarrow \quad & \mathrm{KE}=\dfrac{1}{2} m v^2=\dfrac{n^2 h^2}{8 \pi^2 r^2 m} \quad ……(i) \end{aligned}$
Also, Bohr’s radius for H -atom is, $r=n^2 a_0$
Substituting ’ $r$ ’ in Eq. (i) gives
$\mathrm{KE}=\dfrac{h^2}{8 \pi^2 n^2 a_0^2 m} \text { when } n=2, \mathrm{KE}=\dfrac{h^2}{32 \pi^2 a_0^2 m}$
BETA
