When a proton is accelerated with 1 volt potential difference, then its kinetic energy is

Options

A. 1ev
B. 1840
C. `(1)/(1840)`
D. `931.5 xx 10^(-6)`

Answer

To find the kinetic energy of a proton when it is accelerated through a potential difference of 1 volt, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between potential difference and kinetic energy**:
The kinetic energy (KE) gained by a charged particle when it is accelerated through a potential difference (V) is given by the formula:
\[ KE = q \cdot V \]
where \( q \) is the charge of the particle and \( V \) is the potential difference.

2. **Identify the charge of the proton**:
The charge of a proton is equal in magnitude to the charge of an electron, which is:
\[ q = 1.6 \times 10^{-19} \text{ coulombs} \]

3. **Substitute the values into the formula**:
Given that the potential difference \( V = 1 \text{ volt} \), we can substitute the values into the kinetic energy formula:
\[ KE = (1.6 \times 10^{-19} \text{ C}) \cdot (1 \text{ V}) \]

4. **Calculate the kinetic energy**:
Performing the multiplication gives:
\[ KE = 1.6 \times 10^{-19} \text{ joules} \]

5. **Convert the kinetic energy to electron volts**:
Since 1 electron volt (eV) is defined as the energy gained by a charge of 1 elementary charge (1.6 × 10^-19 C) when accelerated through a potential difference of 1 volt, we can express the kinetic energy in electron volts:
\[ KE = 1 \text{ eV} \]

6. **Conclusion**:
Therefore, the kinetic energy of the proton accelerated through a potential difference of 1 volt is:
\[ KE = 1 \text{ eV} \]

### Final Answer:
The kinetic energy of the proton when accelerated with a 1 volt potential difference is **1 electron volt (1 eV)**.

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