Particle Wavelength Calculation Using de Broglie Relation

This solution details the calculation of the de Broglie wavelength ($\lambda$) for a charged particle accelerated by an electric potential ($V$). The goal is to find the value of $\alpha$ where the wavelength is expressed as $\alpha \times 10^{-12}\text{ m}$.

Given Information

• Particle Charge, $q = 3 \times 10^{-19}\text{ C}$
• Particle Mass, $m = 6 \times 10^{-27}\text{ kg}$
• Accelerating Potential, $V = 1.21\text{ V}$
• Planck's Constant, $h = 6.6 \times 10^{-34}\text{ J.s}$
• Wavelength Format: $\lambda = \alpha \times 10^{-12}\text{ m}$

Step 1: Calculate Kinetic Energy ($KE$)

The kinetic energy gained by the particle is equal to the work done by the electric potential:

$KE = qV$

Substituting the given values:

$KE = (3 \times 10^{-19}\text{ C}) \times (1.21\text{ V}) = 3.63 \times 10^{-19}\text{ J}$

Step 2: Calculate Momentum ($p$)

The relationship between kinetic energy and momentum is:

$KE = \frac{p^2}{2m}$

Solving for momentum:

$p = \sqrt{2m KE}$

Plugging in the values for $m$ and $KE$:

$p = \sqrt{2 \times (6 \times 10^{-27}\text{ kg}) \times (3.63 \times 10^{-19}\text{ J})}$

$p = \sqrt{43.56 \times 10^{-46}\text{ kg}^2\text{m}^2/\text{s}^2}$

$p = 6.6 \times 10^{-23}\text{ kg m/s}$

Step 3: Calculate de Broglie Wavelength ($\lambda$)

Use the de Broglie wavelength formula:

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

Substitute the values of $h$ and $p$:

$\lambda = \frac{6.6 \times 10^{-34}\text{ J.s}}{6.6 \times 10^{-23}\text{ kg m/s}}$

$\lambda = 1 \times 10^{-11}\text{ m}$

Step 4: Determine the Value of $\alpha$

The calculated wavelength is $\lambda = 1 \times 10^{-11}\text{ m}$. We need to express this in the format $\alpha \times 10^{-12}\text{ m}$:

$1 \times 10^{-11}\text{ m} = 10 \times 10^{-12}\text{ m}$

By comparing $\alpha \times 10^{-12}\text{ m}$ with $10 \times 10^{-12}\text{ m}$, we find:

$\alpha = 10$