When a proton is accelerated through 1 V, then its kinetic energy will be:
A) 1840 eV
B) 1 eV
C) 13.6 eV
D) 0.54 eV

To solve this question, we need to understand the relationship between the kinetic energy gained by a charged particle (in this case, a proton) and the potential difference it is accelerated through.

1. **Energy gained by a proton**:
When a proton is accelerated through a potential difference of 1 volt (1 V), it gains an energy equal to the charge of the proton multiplied by the potential difference.

2. **Charge of a proton**:
The charge of a proton is the elementary charge, which is approximately \(1.6 \times 10^{-19}\) coulombs.

3. **Energy gained**:
The energy gained by the proton can be calculated using the formula:
\[
E = q \times V
\]
where \(q\) is the charge of the proton and \(V\) is the potential difference. Since \(q = 1.6 \times 10^{-19}\,C\) and \(V = 1\,V\), we have:
\[
E = (1.6 \times 10^{-19}\,C) \times (1\,V) = 1.6 \times 10^{-19}\,J
\]
The energy in joules can be converted to electron volts (eV) using the conversion factor \(1\,eV = 1.6 \times 10^{-19}\,J\):
\[
E = \frac{1.6 \times 10^{-19}\,J}{1.6 \times 10^{-19}\,J/eV} = 1\,eV
\]

Thus, the kinetic energy gained by the proton is **1 eV**.

**Answer**: B) 1 eV