If an alpha particle and a proton are accelerated from rest by a potential difference of 1MeV, then the ratio of their kinetic energies will be

Options

A. `(1)/(2)`
B. 1
C. 2
D. 4

Answer

To solve the problem of finding the ratio of kinetic energies of an alpha particle and a proton when both are accelerated from rest by a potential difference of 1 MeV, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept**:
When charged particles are accelerated through a potential difference (V), they gain kinetic energy (K.E.) equal to the work done on them by the electric field. This work done can be expressed as:
\[ K.E. = q \cdot V \]
where \( q \) is the charge of the particle and \( V \) is the potential difference.

2. **Identifying the Charges**:
- The charge of a proton (\( q_p \)) is \( +e \) (where \( e \) is the elementary charge).
- The charge of an alpha particle (\( q_{\alpha} \)) is \( +2e \) (since it consists of 2 protons and 2 neutrons).

3. **Calculating Kinetic Energy**:
- For the proton:
\[ K.E._p = q_p \cdot V = e \cdot (1 \text{ MeV}) \]
- For the alpha particle:
\[ K.E._{\alpha} = q_{\alpha} \cdot V = 2e \cdot (1 \text{ MeV}) \]

4. **Finding the Ratio of Kinetic Energies**:
Now we can find the ratio of the kinetic energies of the alpha particle to the proton:
\[ \text{Ratio} = \frac{K.E._{\alpha}}{K.E._p} = \frac{2e \cdot (1 \text{ MeV})}{e \cdot (1 \text{ MeV})} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{2}{1} = 2 \]

5. **Conclusion**:
Therefore, the ratio of the kinetic energies of the alpha particle to the proton is \( 2:1 \).

### Final Answer:
The ratio of the kinetic energies of the alpha particle to the proton is \( 2:1 \).

---