Moving Charges And Magnetism Ques 36
36. An $\alpha$-particle moves in a circular path of radius $0.83$ $ cm$ in the presence of a magnetic field of $0.25 $ $Wb / m^{2}$. The wavelength associated with the particle will be :
[2012]
(a) $1 $ $\AA$
(b) $0.1 $ $\AA$
(c) $10 $ $\AA$
(d) $0.01 $ $\AA$
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Answer:
Correct Answer: 36.(d)
Solution:
- (d) Wavelength
$ 1=\frac{h}{p} \Rightarrow 1=\frac{h}{m v} $
$h=$ plank’s constant $=6.63 \times 10^{-34} J . S$
For circular motion $=F_c=q v B$
$ \begin{aligned} & \Rightarrow \frac{m n^{2}}{r}=q n B \Rightarrow \frac{m v}{q B}=r \\ & r=\frac{m v}{q B} \Rightarrow m v=q r B \end{aligned} $
$ \Rightarrow \quad(2 e)(0.83 \times 10^{-2})(\frac{1}{4}) $
$\lambda=\frac{6.6 \times 10^{-34} \times 4}{2 \times 1.6 \times 10^{-19} \times 0.83 \times 10^{-2}}$
$\lambda=9.93 \times 10^{-34+21} \approx 0.01 $ $\AA$
BETA
