An alpha particle after passing through a potential difference of 3×`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. 9.6×`10^(−13)`J
B. 4.5×`10^(−13)`J
C. 5.6×`10^(−13)`J
D. 1.5×`10^(−13)`J

Explanation

To find the kinetic energy (K.E.) of the alpha particle after passing through a potential difference of \(3 \times 10^6\) volts, we can follow these steps:

### Step 1: Understand the relationship between potential difference and kinetic energy
The kinetic energy gained by a charged particle when it moves through a potential difference \(V\) is given by the formula:
\[ K.E. = q \cdot V \]
where \(q\) is the charge of the particle and \(V\) is the potential difference.

### Step 2: Determine 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:
\[ q = 2 \cdot e \]
where \(e\) (the elementary charge) is approximately \(1.6 \times 10^{-19}\) coulombs. Therefore:
\[ q = 2 \cdot (1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \text{ coulombs} \]

### Step 3: Calculate the kinetic energy in electron volts
Using the potential difference \(V = 3 \times 10^6\) volts, we can calculate the kinetic energy in electron volts:
\[ K.E. = q \cdot V = 2 \cdot (3 \times 10^6) = 6 \times 10^6 \text{ electron volts} \]

### Step 4: Convert kinetic energy from electron volts to joules
To convert the kinetic energy from electron volts to joules, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):
\[ K.E. = 6 \times 10^6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 9.6 \times 10^{-13} \text{ J} \]

### Final Answer
Thus, the kinetic energy of the alpha particle at the time of falling on the foil is:
\[ K.E. = 9.6 \times 10^{-13} \text{ J} \]

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