An `alpha`-particle having kinetic energy 5 MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`- particle reaches is (Atomic no. of Cu=29,`K=9xx10^(9)Nm^(2)//C^(2)`)
(a)`2.35xx10^(-13)` m
(b)`1.67xx10^(-14)` m
(c)`5.98xx10^(-15)` m
(d)none of these

Options

A. (a) `2.35xx10^(-13)` m
B. (b) `1.67xx10^(-14)` m
C. (c) `5.98xx10^(-15)` m
D. (d) none of these

Answer

To solve the problem of finding the shortest distance from the nucleus of copper (Cu) to which an alpha particle can reach, we can use the relationship between kinetic energy and electrostatic potential energy. Here's a step-by-step solution:

### Step 1: Understand the given data
- Kinetic energy (KE) of the alpha particle = 5 MeV
- Atomic number of copper (Cu) = 29
- Coulomb's constant (K) = \(9 \times 10^9 \, \text{N m}^2/\text{C}^2\)

### Step 2: Convert the kinetic energy from MeV to Joules
1 MeV = \(1.6 \times 10^{-13}\) Joules, thus:
\[ KE = 5 \, \text{MeV} = 5 \times 1.6 \times 10^{-13} \, \text{J} = 8.0 \times 10^{-13} \, \text{J} \]

### Step 3: Use the formula relating kinetic energy to the distance
The formula that relates the kinetic energy of the alpha particle to the distance (d) from the nucleus is given by:
\[ KE = \frac{2Z e^2 K}{d} \]
Where:
- \(Z\) = atomic number of copper = 29
- \(e\) = charge of an electron = \(1.6 \times 10^{-19} \, \text{C}\)

### Step 4: Rearrange the formula to solve for d
Rearranging the formula gives:
\[ d = \frac{2Z e^2 K}{KE} \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ d = \frac{2 \times 29 \times (1.6 \times 10^{-19})^2 \times (9 \times 10^9)}{8.0 \times 10^{-13}} \]

### Step 6: Calculate the numerator
Calculating the numerator:
\[ 2 \times 29 = 58 \]
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \]
\[ 58 \times 2.56 \times 10^{-38} \times 9 \times 10^9 = 58 \times 2.56 \times 9 \times 10^{-29} = 1339.2 \times 10^{-29} = 1.3392 \times 10^{-26} \]

### Step 7: Calculate the distance d
Now, substituting back into the equation for d:
\[ d = \frac{1.3392 \times 10^{-26}}{8.0 \times 10^{-13}} = 1.674 \times 10^{-14} \, \text{m} \]

### Step 8: Round off the answer
Rounding off gives:
\[ d \approx 1.67 \times 10^{-14} \, \text{m} \]

### Conclusion
The shortest distance from the nucleus of Cu to which the alpha particle reaches is:
\[ \text{(b) } 1.67 \times 10^{-14} \, \text{m} \]