Q149 / 150 · MHT CET 2009 · Physics

For the Bohr's first orbit of circumference $2 \pi$, the de-Broglie wavelength of revolving electron will be

  • Correct answer
    A
    $2 \pi$
  • B
    $\pi$
  • C
    $\frac{1}{2 \pi r}$
  • D
    $\frac{1}{4 \pi r}$

Solution

$m v r=\frac{n h}{2 \pi}$, according to Bohr's theory $$ \Rightarrow \quad 2 \pi r=n\left(\frac{h}{m v}\right)=n \lambda \quad \text { for } n=1, \lambda=2 \pi r $$

Hence, the correct answer is option: 'A'.

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