Question

An alpha particle moves along a circular path of radius 0.5 mm in a magnetic field of \( 2 \times 10^{-2} \, \text{T} \). The de Broglie wavelength associated with the alpha particle is nearly 
(Planck’s constant \( h = 6.63 \times 10^{-34} \, \text{Js} \))

• A. 3.1 \AA
• B. 1.1 \AA
• C. 0.1 \AA
• D. 2.1 \AA

Hint:

Use \( \lambda = \frac{h}{mv} \), and express \( mv \) in terms of \( qBr \) when a charged particle moves in a circular path under a magnetic field.

Answer 1

The Correct Option is D

Step 1: Use relation for momentum from motion in magnetic field.
\[ r = \frac{mv}{qB} \Rightarrow mv = qBr \] Step 2: Use de Broglie wavelength formula.
\[ \lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{qBr} \] Step 3: Insert known values.
\[ q = 2 \times 1.6 \times 10^{-19} \, \text{C}, \, B = 2 \times 10^{-2} \, \text{T}, \, r = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \] \[ \lambda = \frac{6.63 \times 10^{-34}}{2 \times 1.6 \times 10^{-19} \cdot 2 \times 10^{-2} \cdot 0.5 \times 10^{-3}} = \frac{6.63 \times 10^{-34}}{3.2 \times 10^{-24}} = 2.07 \times 10^{-10} \, \text{m} \] \[ \lambda \approx 2.1 \, \text{\AA} \] Step 4: Select the correct option.
The de Broglie wavelength is approximately 2.1 \AA, which matches option (4).