An a- particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are `lambda_a`
and `lambda_p` respectively. The ratio `(lambda_p)/(lambda_a)`, to the nearest integer, is.

Explanation

To find the ratio of the de Broglie wavelengths of an alpha particle and a proton after being accelerated through a potential difference of 100V, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

### Step 2: Relate kinetic energy to the potential difference
When a charged particle is accelerated through a potential difference \( V \), its kinetic energy (KE) is given by:
\[ KE = QV \]
where \( Q \) is the charge of the particle. For an alpha particle (charge \( Q_\alpha = 2e \)) and a proton (charge \( Q_p = e \)), we can write:
\[ KE_\alpha = 2eV \quad \text{and} \quad KE_p = eV \]

### Step 3: Express velocity in terms of kinetic energy
From the kinetic energy formula, we can express the velocity \( v \) as:
\[ KE = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2KE}{m}} \]
Substituting for kinetic energy, we have:
\[ v = \sqrt{\frac{2QV}{m}} \]

### Step 4: Substitute velocity into the de Broglie wavelength formula
Substituting \( v \) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m \sqrt{\frac{2QV}{m}}} = \frac{h}{\sqrt{2QVm}} \]

### Step 5: Calculate the de Broglie wavelengths for the proton and alpha particle
For the proton:
\[ \lambda_p = \frac{h}{\sqrt{2e \cdot 100 \cdot m_p}} \]
For the alpha particle:
\[ \lambda_\alpha = \frac{h}{\sqrt{2(2e) \cdot 100 \cdot m_\alpha}} = \frac{h}{\sqrt{4e \cdot 100 \cdot m_\alpha}} = \frac{h}{2\sqrt{e \cdot 100 \cdot m_\alpha}} \]

### Step 6: Find the ratio of the wavelengths
Now, we can find the ratio \( \frac{\lambda_p}{\lambda_\alpha} \):
\[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2e \cdot 100 \cdot m_p}}}{\frac{h}{2\sqrt{e \cdot 100 \cdot m_\alpha}}} = \frac{2\sqrt{e \cdot 100 \cdot m_\alpha}}{\sqrt{2e \cdot 100 \cdot m_p}} \]

### Step 7: Simplify the ratio
This simplifies to:
\[ \frac{\lambda_p}{\lambda_\alpha} = 2 \cdot \sqrt{\frac{m_\alpha}{2m_p}} \]
Given that the mass of the alpha particle \( m_\alpha = 4m_p \):
\[ \frac{\lambda_p}{\lambda_\alpha} = 2 \cdot \sqrt{\frac{4m_p}{2m_p}} = 2 \cdot \sqrt{2} = 2\sqrt{2} \]

### Step 8: Calculate the numerical value
Calculating \( 2\sqrt{2} \) gives approximately \( 2.828 \). Rounding to the nearest integer gives:
\[ \frac{\lambda_p}{\lambda_\alpha} \approx 3 \]

### Final Answer
Thus, the ratio \( \frac{\lambda_p}{\lambda_a} \) to the nearest integer is **3**.