Step-by-step Derivation

1. Relating de Broglie Wavelength, Kinetic Energy, and Accelerating Potential 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.

   The kinetic energy ($K$) of a particle with mass $m$ and momentum $p$ is related by $K = \frac{p^2}{2m}$, which can be rearranged to give $p = \sqrt{2mK}$.

   Substituting this expression for momentum into the de Broglie wavelength formula, we get: $\lambda = \frac{h}{\sqrt{2mK}}$

   When a particle with charge $q$ is accelerated from rest through a potential difference $V$, it gains kinetic energy equal to $K = qV$. Substituting this into the wavelength equation gives: $\lambda = \frac{h}{\sqrt{2mqV}}$

2. Properties of the Proton and Alpha-particle We need to identify the mass and charge for both particles.

   • Proton (p):
     • Mass: $m_p$
     • Charge: $q_p = e$ (where $e$ is the elementary charge)
   • Alpha-particle ($\\\\alpha$): An alpha-particle is a helium nucleus (${}^4_2\text{He}$), which consists of 2 protons and 2 neutrons.
     • Mass: $m_\alpha \approx 4 m_p$
     • Charge: $q_\alpha = 2e$

3. Writing Expressions for Wavelengths Using the general formula from Step 1, we can write the de Broglie wavelengths for the proton ($\\\\lambda_p$) and the alpha-particle ($\\\\lambda_\alpha$). Both are accelerated by the same potential difference $V = 100$ V.

   • For the proton: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}} = \frac{h}{\sqrt{2m_p eV}}$

   • For the alpha-particle: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V}} = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16m_p eV}}$

4. Calculating the Ratio {{{\\\lambda _p}} \\over {{\\\\lambda _\\alpha }}} Now we can find the ratio of the two wavelengths: $\frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_p eV}}}{\frac{h}{\sqrt{16m_p eV}}}$

   The constants $h$, $m_p$, $e$, and the potential difference $V$ cancel out, simplifying the expression: $\frac{\lambda_p}{\lambda_\alpha} = \frac{\sqrt{16m_p eV}}{\sqrt{2m_p eV}} = \sqrt{\frac{16}{2}} = \sqrt{8}$

5. Final Numerical Value To find the answer to the nearest integer, we calculate the numerical value of $\\\sqrt{8}$. $\sqrt{8} = 2\sqrt{2} \approx 2 \times 1.4142 = 2.8284$

   Rounding 2.8284 to the nearest integer gives 3.

Final Answer

Thus, the ratio {{{\\\lambda _p}} \\over {{\\\\lambda _\\alpha }}} to the nearest integer is 3.