MHT CET · Physics · Atomic Physics
For the Bohr's first orbit of circumference \(2 \pi\), the de-Broglie wavelength of revolving electron will be
- A \(2 \pi\)
- B \(\pi\)
- C \(\frac{1}{2 \pi r}\)
- D \(\frac{1}{4 \pi r}\)
Answer & Solution
Correct Answer
(A) \(2 \pi\)
Step-by-step Solution
Detailed explanation
\(m v r=\frac{n h}{2 \pi}\), according to Bohr's theory
\(
\Rightarrow 2 \pi r=n\left(\frac{h}{m v}\right)=n \lambda \quad \text { for } n=1, \lambda=2 \pi r
\)
\(
\Rightarrow 2 \pi r=n\left(\frac{h}{m v}\right)=n \lambda \quad \text { for } n=1, \lambda=2 \pi r
\)
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