How much kinetic energy will be gained by an `alpha-`particle in going from a point at `70 V` to another point at `50 V`

Options

A. `40 eV`
B. `40 ke V`
C. `40 MeV`
D. `0eV`

Explanation

To find the kinetic energy gained by an alpha particle when moving from a point at 70 V to another point at 50 V, we can follow these steps:

### Step 1: Understand the concept of potential difference
The kinetic energy gained by a charged particle moving through a potential difference is equal to the work done on the particle by the electric field. This work done can be calculated using the formula:
\[ \text{Work Done} = \text{Charge} \times \text{Potential Difference} \]

### Step 2: Identify the charge of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. The charge of a proton is approximately \( e = 1.6 \times 10^{-19} \) coulombs. Therefore, the total charge \( Q \) of an alpha particle is:
\[ Q = 2 \times e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

### Step 3: Calculate the potential difference
The potential difference \( \Delta V \) that the alpha particle moves through is:
\[ \Delta V = V_{\text{initial}} - V_{\text{final}} = 70 \, \text{V} - 50 \, \text{V} = 20 \, \text{V} \]

### Step 4: Calculate the work done (or kinetic energy gained)
Using the formula for work done:
\[ \text{Work Done} = Q \times \Delta V \]
Substituting the values we have:
\[ \text{Work Done} = (2 \times e) \times (20 \, \text{V}) \]
\[ = 2 \times (1.6 \times 10^{-19} \, \text{C}) \times (20 \, \text{V}) \]
\[ = 2 \times 1.6 \times 20 \times 10^{-19} \]
\[ = 64 \times 10^{-19} \, \text{J} \]
Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), we can convert this to electron volts:
\[ \text{Kinetic Energy} = \frac{64 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 40 \, \text{eV} \]

### Final Answer
The kinetic energy gained by the alpha particle is **40 eV**.

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