A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of `10fm` to the nucleus. The de - Broglie wavelength (in units of fm) of the proton at its start is take the proton mass, `m_p = 5//3xx10^(-27) kg, h//e = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)`.

Options

A. 7
B. 9
C. 11
D. 13

Explanation

To solve the problem step by step, we will follow the principles of conservation of energy and the de Broglie wavelength formula. Here’s how we can approach it:

### Step 1: Identify Given Quantities
- Charge of the nucleus, \( Q = 120e \)
- Charge of the proton, \( q = e \)
- Closest approach distance, \( d = 10 \, \text{fm} = 10 \times 10^{-15} \, \text{m} \)
- Mass of the proton, \( m_p = \frac{5}{3} \times 10^{-27} \, \text{kg} \)
- Planck's constant per charge, \( \frac{h}{e} = 4.2 \times 10^{-15} \, \text{J-s/C} \)
- Coulomb's constant, \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{m/F} \)

### Step 2: Use Conservation of Energy
At the closest approach, all kinetic energy (KE) is converted into potential energy (PE). Thus, we can write:
\[ KE_{\text{initial}} = PE_{\text{final}} \]
Where:
\[ PE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q \cdot q}{d} \]

Substituting the values:
\[ PE = \frac{9 \times 10^9 \cdot (120e) \cdot e}{10 \times 10^{-15}} \]
\[ = \frac{9 \times 10^9 \cdot 120 \cdot (1.6 \times 10^{-19})^2}{10 \times 10^{-15}} \]

### Step 3: Calculate Potential Energy
Calculating \( e^2 \):
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]
Now substituting \( e^2 \) back into the potential energy equation:
\[ PE = \frac{9 \times 10^9 \cdot 120 \cdot 2.56 \times 10^{-38}}{10 \times 10^{-15}} \]
\[ = \frac{9 \times 120 \times 2.56}{10} \times 10^{9 + 15 - 38} \]
\[ = \frac{9 \times 120 \times 2.56}{10} \times 10^{-14} \]

Calculating the numerical values:
\[ = \frac{2764.8}{10} \times 10^{-14} = 276.48 \times 10^{-14} \, \text{J} \]

### Step 4: Relate Kinetic Energy to de Broglie Wavelength
The kinetic energy can also be expressed in terms of the de Broglie wavelength:
\[ \lambda = \frac{h}{p} \quad \text{where } p = \sqrt{2mK} \]
Thus,
\[ \lambda^2 = \frac{h^2}{2mK} \]

### Step 5: Substitute Values
Using \( K = PE \):
\[ \lambda^2 = \frac{h^2}{2m \cdot PE} \]
Substituting \( h = 4.2 \times 10^{-15} \cdot e \):
\[ \lambda^2 = \frac{(4.2 \times 10^{-15} \cdot e)^2}{2 \cdot \frac{5}{3} \times 10^{-27} \cdot 276.48 \times 10^{-14}} \]

### Step 6: Calculate the Wavelength
After substituting the values and simplifying, we can find \( \lambda \) in fm.

### Final Calculation
After performing the calculations, we find:
\[ \lambda \approx 7 \, \text{fm} \]

### Conclusion
The de Broglie wavelength of the proton at its start is approximately \( 7 \, \text{fm} \).