An alpha particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are `lambda_(alpha)` and `lambda_(p)` respectively. The ratio `(lambda_(p))/(lambda_(alpha))` is :

Options

A. `3.8`
B. 8
C. `7.8`
D. `2.8`

Answer

To find the ratio of the de Broglie wavelengths of an alpha particle and a proton after being accelerated through a potential difference of 200 V, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy \( K \) gained by the particle is given by:
\[ K = qV \]
where \( q \) is the charge of the particle.

### Step 2: Determine the charges of the particles
- The charge of a proton \( q_p = e \) (where \( e \) is the elementary charge).
- The charge of an alpha particle \( q_{\alpha} = 2e \) (since it consists of 2 protons and 2 neutrons).

### Step 3: Calculate the kinetic energies
Using the potential difference of 200 V:
- For the proton:
\[ K_p = q_p V = e \cdot 200 = 200e \text{ Joules} \]
- For the alpha particle:
\[ K_{\alpha} = q_{\alpha} V = 2e \cdot 200 = 400e \text{ Joules} \]

### Step 4: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ K = \frac{p^2}{2m} \]
From this, we can express momentum as:
\[ p = \sqrt{2mK} \]

### Step 5: Calculate the momenta of the particles
- For the proton:
\[ p_p = \sqrt{2m_p K_p} = \sqrt{2m_p \cdot 200e} \]
- For the alpha particle:
\[ p_{\alpha} = \sqrt{2m_{\alpha} K_{\alpha}} = \sqrt{2m_{\alpha} \cdot 400e} \]

### Step 6: Calculate the de Broglie wavelengths
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

- For the proton:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2m_p \cdot 200e}} \]
- For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\sqrt{2m_{\alpha} \cdot 400e}} \]

### Step 7: Find the ratio of the wavelengths
Now, we can find the ratio \( \frac{\lambda_p}{\lambda_{\alpha}} \):
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_p \cdot 200e}}}{\frac{h}{\sqrt{2m_{\alpha} \cdot 400e}}} = \frac{\sqrt{2m_{\alpha} \cdot 400e}}{\sqrt{2m_p \cdot 200e}} \]
This simplifies to:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha} \cdot 400}}{\sqrt{m_p \cdot 200}} = \sqrt{\frac{m_{\alpha} \cdot 400}{m_p \cdot 200}} = \sqrt{2 \cdot \frac{m_{\alpha}}{m_p}} \]

### Step 8: Substitute the masses
The mass of an alpha particle \( m_{\alpha} \) is approximately 4 times the mass of a proton \( m_p \):
\[ \frac{m_{\alpha}}{m_p} \approx 4 \]
Thus:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{2 \cdot 4} = \sqrt{8} = 2\sqrt{2} \]

### Final Answer
The ratio \( \frac{\lambda_p}{\lambda_{\alpha}} \) is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = 2\sqrt{2} \]