De-Broglie Wavelength and Accelerating Potential Explanation

This problem involves understanding the relationship between the de-Broglie wavelength of a particle and the electric potential it moves through. We need to find the potential required for an alpha particle to have the same de-Broglie wavelength as a proton accelerated by a potential V.

De-Broglie Wavelength Formula

The de-Broglie wavelength ($\lambda$) of a particle is given by the formula:

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

where h is Planck's constant and p is the momentum of the particle.

Relating Wavelength to Accelerating Potential

When a particle with charge q is accelerated through a potential difference V, it gains kinetic energy (KE) equal to:

$ KE = qV $

The kinetic energy is also related to momentum (p) and mass (m) by:

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

Equating these two expressions for kinetic energy:

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

Solving for momentum, p:

$ p = \sqrt{2mqV} $

Now, substitute this expression for momentum back into the de-Broglie wavelength formula:

$ \lambda = \frac{h}{\sqrt{2mqV}} $

This equation shows that the de-Broglie wavelength is inversely proportional to the square root of the product of mass, charge, and accelerating potential.

Comparing Proton and Alpha Particle

Let's compare the properties of a proton and an alpha particle:

• Proton:
  • Charge: $q_p = e$ (elementary charge)
  • Mass: $m_p$
  • Accelerating Potential: $V_p = V$
  • De-Broglie Wavelength: $ \lambda_p = \frac{h}{\sqrt{2 m_p e V}} $
• Alpha Particle (Helium Nucleus):
  • Charge: $q_\alpha = 2e$ (contains 2 protons)
  • Mass: $m_\alpha \approx 4 m_p$ (approximately 4 times the mass of a proton)
  • Accelerating Potential: $V_\alpha$ (this is what we need to find)
  • De-Broglie Wavelength: $ \lambda_\alpha = \frac{h}{\sqrt{2 m_\alpha q_\alpha V_\alpha}} $

Calculating the Required Potential

The question states that the de-Broglie wavelength must be the same for both particles:

$ \lambda_\alpha = \lambda_p $

Substituting the expressions for the wavelengths:

$ \frac{h}{\sqrt{2 m_\alpha q_\alpha V_\alpha}} = \frac{h}{\sqrt{2 m_p e V}} $

We can cancel out Planck's constant (h) and the factor of 2 from both sides:

$ \sqrt{m_\alpha q_\alpha V_\alpha} = \sqrt{m_p e V} $

Squaring both sides gives:

$ m_\alpha q_\alpha V_\alpha = m_p e V $

Now, substitute the properties of the alpha particle ($m_\alpha = 4 m_p$ and $q_\alpha = 2e$):

$ (4 m_p) (2e) V_\alpha = m_p e V $

$ 8 m_p e V_\alpha = m_p e V $

Cancel out $m_p$ and $e$ from both sides:

$ 8 V_\alpha = V $

Solve for $V_\alpha$:

$ V_\alpha = \frac{V}{8} $

Conclusion

Therefore, the required accelerating potential for the alpha particle to have the same de-Broglie wavelength as the proton is V/8.