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Physics Question on de broglie hypothesis

An α\alpha-particle moves in a circular path of radius 1 cm1\, cm in a uniform magnetic field of 0.125 T0.125\, T. The de Broglie wavelength associated with the α\alpha-particle is

A

1.65×10−12m1.65 \times 10^{-12} m

B

3.3×10−12m3.3 \times 10^{-12} m

C

4.95×10−12m4.95 \times 10^{-12} m

D

6.6×10−12m6.6 \times 10^{-12} m

Answer

1.65×10−12m1.65 \times 10^{-12} m

Explanation

Solution

Given magnetic field, B=0.125 TB=0.125 \,T,
Radius of the circular path, R=1 cm=10−2mR=1 \,cm =10^{-2} m
We know radius of circular path in a uniform magnetic field,
R=mvqB⇒mv=qBRR=\frac{m v}{q B} \Rightarrow m v=q B R
and de -Broglie wavelength, λ=hmv\lambda=\frac{h}{m v}
[∵q=qα\left[\because q=q_{\alpha}\right., For α\alpha - particle ]]
or λ=hqαBR\lambda=\frac{h}{q_{\alpha} B R}
λ=6.6×10−342×1.6×10−19×0.125×1×10−2\lambda=\frac{6.6 \times 10^{-34}}{2 \times 1.6 \times 10^{-19} \times 0.125 \times 1 \times 10^{-2}}
[∵qα=+2C=2×1.6×10−19]{\left[\because q_{\alpha}=+2 C =2 \times 1.6 \times 10^{-19}\right]}
⇒λ=1.65×10−12m\Rightarrow \lambda=1.65 \times 10^{-12} m
Hence, de-Broglie wavelength associated with α\alpha -particle is 1.65×1012m1.65 \times 10^{12} m.