An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². The de-Broglie wavelength associated with the particle will be?

Steps to Solve

1. Convert Units

   Convert the radius from centimeters to meters. Since (1 \text{ cm} = 0.01 \text{ m}):

   $$ r = 0.83 \text{ cm} \times 0.01 \text{ m/cm} = 0.0083 \text{ m} $$

2. Calculate the Momentum

   Use the formula for the momentum (p) of a charged particle in a magnetic field:

   $$ p = q \cdot B \cdot r $$

   For an α-particle, (q = 2e) (where (e) is the elementary charge, (1.6 \times 10^{-19} \text{ C})), so:

   $$ q = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C} $$

   Now substituting (q), (B = 0.25 \text{ Wb/m}^2), and (r = 0.0083 \text{ m}):

   $$ p = (3.2 \times 10^{-19}) \cdot (0.25) \cdot (0.0083) $$

3. Calculate the de Broglie Wavelength

   The de Broglie wavelength (\lambda) is given by:

   $$ \lambda = \frac{h}{p} $$

   Where (h) is Planck's constant, (h = 6.63 \times 10^{-34} \text{ Js}).

   Substitute the value of (p) calculated in the previous step into the equation.

4. Final Calculation

   After calculating (p), compute (\lambda) using:

   $$ \lambda = \frac{6.63 \times 10^{-34}}{p} $$