An alpha particle after passing through a potential difference of 5×`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. 12.8×`10^(−13)`J
B. 16×`10^(−13)`J
C. 18×`10^(−13)`J
D. 14×`10^(−13)`J

Explanation

To find the kinetic energy (K.E.) of an alpha particle after passing through a potential difference of \(5 \times 10^6\) volts, we can follow these steps:

### Step 1: Understand the charge of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. Therefore, it has a charge of \(+2e\), where \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) coulombs).

### Step 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 the formula:
\[ \text{K.E.} = qV \]
where \(q\) is the charge of the particle and \(V\) is the potential difference.

### Step 3: Substitute the values
For an alpha particle:
- Charge \(q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C}\)
- Potential difference \(V = 5 \times 10^6 \, \text{V}\)

Substituting these values into the formula:
\[ \text{K.E.} = (2 \times 1.6 \times 10^{-19}) \times (5 \times 10^6) \]

### Step 4: Calculate the kinetic energy in electron volts
Calculating the above expression:
\[ \text{K.E.} = 3.2 \times 10^{-19} \times 5 \times 10^6 = 16 \times 10^{-13} \, \text{J} \]

### Step 5: Convert to electron volts
Since \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\), we can convert the kinetic energy from joules to electron volts:
\[ \text{K.E.} = \frac{16 \times 10^{-13}}{1.6 \times 10^{-19}} = 10^7 \, \text{eV} \]

### Conclusion
Thus, the kinetic energy of the alpha particle at the time of falling on the foil is:
\[ \text{K.E.} = 10^7 \, \text{eV} \]