This solution explains how to find the ratio of the de Broglie wavelength of a proton to that of an alpha particle when they are accelerated through different potential differences resulting in the same final kinetic energy.

De Broglie Wavelength Fundamentals

The de Broglie wavelength ($\lambda$) of a particle is defined as the wavelength associated with its momentum ($p$). The formula connecting wavelength, Planck's constant ($h$), and momentum is:

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

This equation is fundamental in understanding wave-particle duality.

Relating Wavelength, Momentum, and Kinetic Energy

The kinetic energy ($KE$) of a particle depends on its mass ($m$) and momentum ($p$) through the equation:

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

By rearranging this formula, we can express the momentum of a particle in terms of its mass and kinetic energy:

$ p = \sqrt{2m \cdot KE} $

Now, substitute this expression for momentum ($p$) back into the de Broglie wavelength formula:

$ \lambda = \frac{h}{\sqrt{2m \cdot KE}} $

This formula shows that for a given kinetic energy, the wavelength is inversely proportional to the square root of the mass.

Proton and Alpha Particle Properties

The problem provides specific conditions relating the proton and the alpha particle:

• Identical Kinetic Energies: Both the proton and the alpha particle have the same final kinetic energy. We can represent this as $KE_p = KE_\alpha$. Let's call this common value $KE$.
• Mass Relationship: The mass of an alpha particle ($m_\alpha$) is given as approximately four times the mass of a proton ($m_p$). Mathematically, this is $m_\alpha \approx 4m_p$.

Step-by-Step Ratio Calculation

We need to find the ratio of the de Broglie wavelength of the proton ($\lambda_p$) to that of the alpha particle ($\lambda_\alpha$). Let's write the expressions for each:

• For the proton: $ \lambda_p = \frac{h}{\sqrt{2m_p \cdot KE_p}} $ Since $KE_p = KE$, this becomes: $ \lambda_p = \frac{h}{\sqrt{2m_p \cdot KE}} $
• For the alpha particle: $ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha \cdot KE_\alpha}} $ Using $KE_\alpha = KE$ and $m_\alpha = 4m_p$, this becomes: $ \lambda_\alpha = \frac{h}{\sqrt{2(4m_p) \cdot KE}} = \frac{h}{\sqrt{8m_p \cdot KE}} $

Now, let's calculate the ratio $\frac{\lambda_p}{\lambda_\alpha}$:

$ \frac{\lambda_p}{\lambda_\alpha} = \frac{\left(\frac{h}{\sqrt{2m_p \cdot KE}}\right)}{\left(\frac{h}{\sqrt{8m_p \cdot KE}}\right)} $

Simplify the expression by canceling Planck's constant ($h$) and inverting the denominator:

$ \frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{8m_p \cdot KE}}{\sqrt{2m_p \cdot KE}} $

Combine the terms under a single square root:

$ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{8m_p \cdot KE}{2m_p \cdot KE}} $

Cancel out the common factors $m_p$ and $KE$:

$ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{8}{2}} $

Calculate the final value:

$ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{4} = 2 $

Final Wavelength Ratio Conclusion

The calculation shows that the ratio of the de Broglie wavelength of the proton to that of the alpha particle is 2.

Therefore, the ratio $\lambda_p : \lambda_\alpha$ is $2 : 1$.