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. $1\, ?$
• B. $0.1 \, ?$
• C. $10\, ?$
• D. $0.01\, ?$

Answer 1

The Correct Option is D

To determine the de Broglie wavelength associated with an alpha particle moving in a circular path within a magnetic field, we are given the following parameters:

• The radius of the circular path, r = 0.83 \, \text{cm} = 0.0083 \, \text{m}
• The magnetic field, B = 0.25 \, \text{Wb/m}^2
• Planck's constant, h = 6.63 \times 10^{-34} \, \text{J s}
• Elementary charge, e = 1.6 \times 10^{-19} \, \text{C}

An alpha particle is a helium nucleus consisting of 2 protons and 2 neutrons. Its charge is 2e and its mass is approximately 4 \times \text{mass of proton} = 4 \times 1.67 \times 10^{-27} \, \text{kg}.

First, calculate the momentum of the alpha particle in the magnetic field using the formula:

p = qBr

• q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C}
• Thus, p = 3.2 \times 10^{-19} \times 0.25 \times 0.0083 \, \text{kg m/s}

Computing the above:

p = 6.64 \times 10^{-21} \, \text{kg m/s}

Now, using the de Broglie wavelength formula:

\lambda = \frac{h}{p}

Substitute the known values:

\lambda = \frac{6.63 \times 10^{-34}}{6.64 \times 10^{-21}}

Calculating this gives:

\lambda \approx 0.01 \, \text{m}

Therefore, the de Broglie wavelength associated with the alpha particle is 0.01 \, \text{m}.