A proton is about `1840` times heavier than an electron. When it is accelerated by a potential difference of `1 kV`. Its kinetic energy will be

Options

A. `1840 keV`
B. `1//1840 keV`
C. `1 keV`
D. `920 keV`

Explanation

To solve the problem of finding the kinetic energy of a proton accelerated by a potential difference of 1 kV, we can follow these steps:

### Step 1: Understand the relationship between potential difference and kinetic energy
When a charged particle is accelerated through a potential difference (V), the kinetic energy (KE) gained by the particle is given by the equation:
\[ KE = q \cdot V \]
where \( q \) is the charge of the particle and \( V \) is the potential difference.

### Step 2: Identify the charge of the proton
The charge of a proton is equal to the elementary charge \( e \), which is approximately:
\[ e \approx 1.6 \times 10^{-19} \text{ C} \]

### Step 3: Substitute the values into the kinetic energy equation
Given that the potential difference \( V = 1 \text{ kV} = 1000 \text{ V} \), we can substitute the values into the kinetic energy equation:
\[ KE = e \cdot V = (1.6 \times 10^{-19} \text{ C}) \cdot (1000 \text{ V}) \]

### Step 4: Calculate the kinetic energy
Now, we perform the multiplication:
\[ KE = 1.6 \times 10^{-19} \text{ C} \cdot 1000 \text{ V} = 1.6 \times 10^{-16} \text{ J} \]

### Step 5: Convert the kinetic energy to electron volts
Since the problem may require the answer in electron volts (eV), we can convert joules to electron volts using the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ KE = \frac{1.6 \times 10^{-16} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 1000 \text{ eV} = 1 \text{ keV} \]

### Final Answer
Thus, the kinetic energy of the proton when accelerated by a potential difference of 1 kV is:
\[ KE = 1 \text{ keV} \]