An alpha particle after passing through a potential difference of 1×`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. 1.2×`10^(−13)`J
B. 3.2×`10^(−13)`J
C. 12.8×`10^(−13)`J
D. None of these

Explanation

To find the kinetic energy (K.E.) of an alpha particle after passing through a potential difference of \(1 \times 10^6\) 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. Since it has 2 protons, the charge of the alpha particle is:
\[ q = 2e \]
where \(e\) (the charge of an electron) is approximately \(1.6 \times 10^{-19}\) coulombs.

### Step 2: Calculate the Kinetic Energy in Electron Volts
The kinetic energy gained by a charged particle when it moves through a potential difference \(V\) is given by:
\[ \text{K.E.} = qV \]
Substituting the values:
\[ \text{K.E.} = (2e)(1 \times 10^6 \text{ V}) = 2 \times (1.6 \times 10^{-19} \text{ C})(1 \times 10^6 \text{ V}) \]
Calculating this gives:
\[ \text{K.E.} = 2 \times 1.6 \times 10^{-19} \times 10^6 = 3.2 \times 10^{-13} \text{ joules} \]

### Step 3: Convert Kinetic Energy to Joules
Since we already calculated the kinetic energy in joules in the previous step, we have:
\[ \text{K.E.} = 3.2 \times 10^{-13} \text{ joules} \]

### Conclusion
The kinetic energy of the alpha particle at the time of falling on the silver foil is:
\[ \text{K.E.} = 3.2 \times 10^{-13} \text{ joules} \]