The ratio of moment of an electron and an `alpha`-particle which are accelerated from rest by a potential difference of `100V` is

Options

A. 1
B. `sqrt((2m_(e))/(m_(alpha)))`
C. `sqrt((m_(e))/(m_(alpha)))`
D. `sqrt((m_(e))/(2m_(alpha)))`

Answer

To solve the problem of finding the ratio of the momenta of an electron and an alpha particle accelerated from rest by a potential difference of 100V, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential energy
When a charged particle is accelerated through a potential difference \( V \), the kinetic energy (K.E.) gained by the particle is equal to the work done on it by the electric field. This can be expressed as:
\[ K.E. = Q \cdot V \]
where \( Q \) is the charge of the particle.

### Step 2: Write the expression for momentum
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ K.E. = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle.

### Step 3: Equate the two expressions for kinetic energy
From the two equations above, we can set them equal to each other:
\[ \frac{p^2}{2m} = Q \cdot V \]

### Step 4: Solve for momentum \( p \)
Rearranging the equation gives:
\[ p^2 = 2m \cdot Q \cdot V \]
Taking the square root, we find:
\[ p = \sqrt{2m \cdot Q \cdot V} \]

### Step 5: Calculate the momenta for the electron and the alpha particle
For the electron:
- Charge \( Q_e = e \) (where \( e \) is the elementary charge)
- Mass \( m_e \)

Thus, the momentum of the electron \( p_e \) is:
\[ p_e = \sqrt{2m_e \cdot e \cdot V} \]

For the alpha particle:
- Charge \( Q_{\alpha} = 2e \) (since it has two protons)
- Mass \( m_{\alpha} \) (approximately 4 times the mass of the proton)

Thus, the momentum of the alpha particle \( p_{\alpha} \) is:
\[ p_{\alpha} = \sqrt{2m_{\alpha} \cdot 2e \cdot V} \]

### Step 6: Find the ratio of momenta
Now, we will find the ratio of the momenta:
\[ \frac{p_e}{p_{\alpha}} = \frac{\sqrt{2m_e \cdot e \cdot V}}{\sqrt{2m_{\alpha} \cdot 2e \cdot V}} \]

### Step 7: Simplify the ratio
The terms \( 2 \), \( e \), and \( V \) cancel out:
\[ \frac{p_e}{p_{\alpha}} = \frac{\sqrt{m_e}}{\sqrt{2m_{\alpha}}} \]

### Final Answer
Thus, the ratio of the momenta of the electron to the alpha particle is:
\[ \frac{p_e}{p_{\alpha}} = \sqrt{\frac{m_e}{2m_{\alpha}}} \]