Q151 / 200 · NEET 2012 (Screening) · Physics

An $\alpha$-particle moves in a circular path of radius $0.83 \mathrm{~cm}$ in the presence of a magnetic field of $0.25 \mathrm{~Wb} / \mathrm{m}^2$. The de-Broglie wavelength associated with the particle will be

  • A
    $1 Å$
  • B
    $0.1 Å$
  • C
    $10 Å$
  • Correct answer
    D
    $0.01 Å$

Solution

We knows $R=\frac{m v}{q B}$ $\begin{aligned} \lambda & =\frac{h}{m v} \\ \lambda & =\frac{h}{q B R} \\ & =\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19} \times 0.83 \times 10^{-2} \times 0.25} \\ & =0.01 Å \end{aligned}$

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

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