This question involves finding the ratio of de Broglie wavelengths for an α-particle and a proton after both are accelerated through the same potential difference. Let's break this down step by step.

Step 1: Recall the de Broglie wavelength formula
The de Broglie wavelength (λ) for a particle is given by: $\lambda =\frac{h}{p}$ where h is Planck's constant and p is the momentum of the particle.

Step 2: Relate momentum to kinetic energy
The kinetic energy (K) of a particle is related to its momentum by: $K=\frac{p^2}{2m}$ where m is the mass of the particle. We can rearrange this to express momentum in terms of kinetic energy: $p=\sqrt{2mK}$ Substituting this into the de Broglie formula gives: $\lambda =\frac{h}{\sqrt{2mK}}$

Step 3: Find the kinetic energy gained from the potential difference
When a charged particle is accelerated from rest by a potential difference (V), it gains a kinetic energy equal to the work done by the electric field. This is given by: $K=qV$ where q is the charge of the particle. Since both particles start from rest and are accelerated by the same potential difference (V = 100 V), their kinetic energies are: $K_p=q_{p}V\text{and}{K}_{\alpha }=q_{\alpha }V$

Step 4: Write the de Broglie wavelength for each particle
Using the formula from Step 2 and substituting K = qV, we get the wavelength for a particle: $\lambda =\frac{h}{\sqrt{2mqV}}$ Therefore, the wavelengths for the proton (λ_p) and the α-particle (λ_α) are: ${\lambda }_p=\frac{h}{\sqrt{2m_p{q}_{p}V}}\text{and}{\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }V}}$

Step 5: Find the ratio of the wavelengths
We are asked to find the ratio λ_p/λ_α. $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{\frac{h}{\sqrt{2m_p{q}_{p}V}}}{\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }V}}}=\frac{\sqrt{2m_{\alpha }{q}_{\alpha }V}}{\sqrt{2m_p{q}_{p}V}}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}$ Notice that Planck's constant (h) and the potential difference (V) cancel out. The ratio depends only on the masses and charges of the particles.

Step 6: Substitute the known values
An α-particle is a helium nucleus.

• Mass of proton, m_p
• Mass of α-particle, m_α = 4 × m_p
• Charge of proton, q_p = e (elementary charge)
• Charge of α-particle, q_α = 2e

Substituting these values into our ratio: $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}=\sqrt{\frac{\left(4m_p\right)\left(2e\right)}{m_{p}e}}=\sqrt{\frac{8}{1}}=\sqrt{8}=2\sqrt{2}$

Step 7: Find the numerical value to the nearest integer
$2\sqrt{2}\approx 2\times 1.414\approx 2.828$ The nearest integer to 2.828 is 3.

Final Answer: The ratio λ_p/λ_α, to the nearest integer, is 3.

Related Concepts and Formulae

De Broglie Hypothesis: Proposed by Louis de Broglie, it states that all matter exhibits wave-like properties. The wavelength associated with a particle is inversely proportional to its momentum.

Key Formula: $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$

Acceleration of Charged Particles: When a charged particle is accelerated from rest through a potential difference V, it gains a kinetic energy K = qV. This is a fundamental concept in particle physics and electronics (e.g., electron guns).