Question

An α-particle moves in a circular path of radius $0.83cm$ in the presence of a magnetic field of $0.25Wb/m^2$. The de Broglie wavelength associated with the particle will be:

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

Answer 1

The correct option is D

$0.01\dot{A}$

Step 1: Given

Charge of alpha particle: $q=2e=2\times 1.6\times {10}^{-19}$

Radius of circular path: $R=0.83cm=0.83\times {10}^{-2}m$

Strength of magnetic field: $B=0.25Wb/m^2$

Step 2: Formula Used

$R=\frac{mv}{Bq}$

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

Step 3: Find the de Broglie wavelength

Calculate the value of the product of mass and velocity

$$\begin{gathered} R=\frac{mv}{Bq} \\ \Rightarrow mv=RBq \\ \Rightarrow mv=0.83m\times {10}^{-2}\times 0.25\times 2\times 1.6\times {10}^{-19} \end{gathered}$$

Calculate the de Broglie wavelength by using the formula $\lambda =\frac{h}{mv}$ and substituting the values

$$\begin{gathered} \lambda =\frac{6.6\times {10}^{-34}}{0.83\times {10}^{-2}\times 0.25\times 2\times 1.6\times {10}^{-19}} \\ =\frac{6.6\times {10}^{-34}}{6.6\times {10}^{-22}} \\ =1\times {10}^{-12} \\ =0.01\dot{A} \end{gathered}$$

Hence, Option (D) is correct. the de Broglie wavelength is $0.01\dot{A}$.