AP EAMCET · PHYSICS · Atomic Physics
In Bohr's theory the potential of an electron at a position is \(\frac{K r^2}{2}\), where \(K\) is a constant. Then, the quantised energy of the electron in \(n\)th orbit is
- A \(\frac{n h K}{2 \pi m}\)
- B \(\frac{n h}{2 \pi} \sqrt{\frac{K}{m}}\)
- C \(\frac{n h m}{2 \pi K}\)
- D \(\frac{n h}{2 \pi} \sqrt{\frac{m}{K}}\)
Answer & Solution
Correct Answer
(B) \(\frac{n h}{2 \pi} \sqrt{\frac{K}{m}}\)
Step-by-step Solution
Detailed explanation
Given, potential energy of electron, \(\begin{aligned} & & U=\frac{K r^2}{2} \\ \therefore & F & =\frac{d U}{d r}=\frac{d}{d r} \frac{K r^2}{2}=K r \\ \text {But } & F & =\frac{m v^2}{r} \Rightarrow K r=\frac{m v^2}{r} \\ \Rightarrow & v & =r \sqrt{\frac{K}{m}} \end{aligned}\)…
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