Question

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 de Broglie wavelength associated with the particle will be
(Take $h = 6.63 × 10^{-34} J\, s$, $e =1.6 ×10^{-19}C$)

• A. $1\,\mathring{A}$
• B. $0.1 \,\mathring{A}$
• C. $10\,\mathring{A}$
• D. $0.01\,\mathring{A}$

Answer 1

Answer: (D)

Radius of the circular path of a charged particle in a magnetic field is given by
$R= \frac {mv}{Bq} $ or $ mv=RBq$
Here, $R=0.83\, cm=0.83 \times 10 ^{-2}m$
$B=0.25 \, Wb \, m^2$
$q=2e=2 \times 1.6 \times 10^{-19}C$
$\therefore mv=(0.83 \times 10^{-2})(0.25)(2 \times 1.6 \times 10^{-19})$
de Broglie wavelength,
$\lambda=\frac {h}{mv}= \frac {6.6 \times 10^{-34}}{0.83 \times 10^{-2}\times 0.25 \times 2 \times 1.6 \times 10^{-19}}$
$=0.01 \,\mathring{A}$