A proton, a deutron and an `alpha`- particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to it. What is the ratio of their kinetic energy?

Options

A. `1:1:2`
B. `2:2:1`
C. `1:2:1`
D. `2:1:1`

Answer

To solve the problem of finding the ratio of kinetic energy of a proton, a deuteron, and an alpha particle accelerated through the same potential difference and entering a uniform magnetic field, 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 \( KE \) gained by the particle is given by the equation:
\[ KE = qV \]
where \( q \) is the charge of the particle.

### Step 2: Identify the charges of the particles
- The charge of a proton \( (q_p) \) is \( +e \).
- The charge of a deuteron \( (q_d) \) is also \( +e \) (since it has one proton and one neutron).
- The charge of an alpha particle \( (q_{\alpha}) \) is \( +2e \) (since it consists of two protons and two neutrons).

### Step 3: Write the kinetic energy expressions for each particle
Using the relationship from Step 1, we can express the kinetic energies of each particle:
- For the proton:
\[ KE_p = q_p V = eV \]
- For the deuteron:
\[ KE_d = q_d V = eV \]
- For the alpha particle:
\[ KE_{\alpha} = q_{\alpha} V = 2eV \]

### Step 4: Find the ratio of their kinetic energies
Now, we can find the ratio of the kinetic energies of the three particles:
\[ \text{Ratio} = KE_p : KE_d : KE_{\alpha} = eV : eV : 2eV \]
This simplifies to:
\[ 1 : 1 : 2 \]

### Conclusion
Thus, the ratio of the kinetic energies of the proton, deuteron, and alpha particle is:
\[ 1 : 1 : 2 \]