An \( \alpha \)-particle moves in a circular paths of radius 0.83 cm in the presence of a magnetic field \( 0.25\,\mathrm{Wb/m^2} \). The de-Broglie wavelength associated with the particle would be approximately:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.01 Å
The magnetic force supplies the centripetal force, \( qvB = mv^{2}/r \), so the momentum is \( p = mv = qBr \). The de Broglie wavelength is \( \lambda = h/p \).
The α-particle has charge \( q = 2e = 3.2\times10^{-19}\,\mathrm{C} \) and \( r = 0.83\,\mathrm{cm} = 0.83\times10^{-2}\,\mathrm{m} \).
\( p = qBr = 3.2\times10^{-19}\times0.25\times0.83\times10^{-2} \approx 6.64\times10^{-22}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle \lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{6.64\times10^{-22}} \approx 1\times10^{-12}\,\mathrm{m} \)
Since \( 1\,\mathrm{\mathring{A}} = 10^{-10}\,\mathrm{m} \), \( \lambda = 0.01\,\mathrm{\mathring{A}} \).