An alpha particle after passing through a potential difference of 2×`10^6` volt falls on a silver foil. The atomic number of silver is 47. The K.E. of the α-particle at the time of falling on the foil is:

Options

A. 4.5×`10^(−13)`J
B. 6.4×`10^(−13)`J
C. 7.8×`10^(−13)`J
D. 5.8×`10^(−13)`J

Explanation

To find the kinetic energy (K.E.) of an alpha particle after it has passed through a potential difference of \(2 \times 10^6\) volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Charge of the Alpha Particle:**
An alpha particle consists of 2 protons and 2 neutrons. Since neutrons do not carry charge, the charge of the alpha particle is due to the protons only. The charge of a proton is approximately \(1.6 \times 10^{-19}\) coulombs. Therefore, the total charge \(Q\) of the alpha particle is:
\[ Q = 2 \times (1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \text{ C} \]

2. **Use the Formula for Kinetic Energy:**
The kinetic energy gained by a charged particle when it moves through a potential difference \(V\) is given by:
\[ \text{K.E.} = Q \times V \]
where \(Q\) is the charge of the particle and \(V\) is the potential difference.

3. **Substitute the Values:**
Here, the potential difference \(V\) is \(2 \times 10^6\) volts. Substituting the values into the formula:
\[ \text{K.E.} = (3.2 \times 10^{-19} \text{ C}) \times (2 \times 10^6 \text{ V}) \]

4. **Calculate the Kinetic Energy:**
\[ \text{K.E.} = 3.2 \times 2 \times 10^{-19} \times 10^6 = 6.4 \times 10^{-13} \text{ J} \]

5. **Final Result:**
The kinetic energy of the alpha particle at the time of falling on the foil is:
\[ \text{K.E.} = 6.4 \times 10^{-13} \text{ J} \]