Step-by-step Derivation

1. Analyze the Physical Scenario and Apply Conservation of Energy

   A proton with charge e is fired from a very large distance towards a nucleus with charge Q = 120e. Initially, the proton has kinetic energy $K_i$ and its potential energy $U_i$ is negligible, so $U_i ≈ 0$. The total initial energy is $E_i = K_i$.

   The proton travels towards the nucleus and its kinetic energy is converted into electrostatic potential energy due to repulsion. At the point of closest approach, $r_min$, the proton momentarily stops before being repelled. At this point, its kinetic energy $K_f = 0$. The potential energy $U_f$ is given by Coulomb's law: $U_f = \frac{1}{4\pi \varepsilon_0} \frac{(e)(Q)}{r_{min}} = \frac{k(e)(120e)}{r_{min}} = \frac{120 k e^2}{r_{min}}$

   According to the principle of conservation of energy, the total initial energy equals the total final energy: $E_i = E_f$ $K_i = U_f = \frac{120 k e^2}{r_{min}}$

2. Relate Kinetic Energy to de Broglie Wavelength

   The de Broglie wavelength λ of a particle is related to its momentum p by the equation λ = h/p. The kinetic energy K is related to momentum p and mass $m_p$ by $K = p^2 / (2m_p)$. Therefore, $p = \sqrt{2m_p K}$.

   Substituting this into the de Broglie wavelength formula, we get the initial wavelength λ of the proton: $\lambda = \frac{h}{\sqrt{2m_p K_i}}$

3. Combine Formulas and Solve for Wavelength

   Substitute the expression for $K_i$ from Step 1 into the wavelength equation: $\lambda = \frac{h}{\sqrt{2m_p \left( \frac{120 k e^2}{r_{min}} \right)}} = \frac{h}{\sqrt{\frac{240 m_p k e^2}{r_{min}}}}$

   To make use of the given constant h/e, we can rearrange the expression: $\lambda = \frac{h/e}{\sqrt{\frac{240 m_p k}{r_{min}}}}$

4. Address the Proton Mass Value and Perform Calculations

   The problem states the proton mass as $m_p = (5 × 3) × 10^{−27}$ kg, which is $15 × 10^{−27}$ kg. This value is nearly 9 times the actual proton mass and leads to a non-integer answer, which is unlikely for an integer-type question. It is highly probable that this is a typo and the intended value was $m_p = (5/3) × 10^{−27}$ kg, which is a common approximation for the proton mass ($1.67 × 10^{−27}$ kg). Using this corrected value leads to a clean integer answer.

   Let's proceed with the calculation using $m_p = (5/3) × 10^{−27}$ kg.

   Given values:

   • $r_{min} = 10$ fm $= 10 × 10^{−15}$ m
   • $m_p = (5/3) × 10^{−27}$ kg
   • $h/e = 4.2 × 10^{−15}$ J.s/C
   • $k = 1 / (4πε_0) = 9 × 10^9$ N m²/C²

   First, calculate the term under the square root: $\frac{240 m_p k}{r_{min}} = \frac{240 \times (\frac{5}{3} \times 10^{-27}) \times (9 \times 10^9)}{10 \times 10^{-15}}$ $= \frac{240 \times 5 \times 9}{3 \times 10} \times 10^{-27+9+15} = (80 \times 5 \times 3) \times 10^{-3} = 1200 \times 10^{-3} = 1.2$ Let me re-calculate that part: $= \frac{240 \times 5 \times 9}{3 \times 10} \times 10^{-3} = (80 \times 5 \times 3 / 10) imes 10^{-3} = (24 \times 5 \times 3) \times 10^{-3} = 360 \times 10^{-3} = 0.36$

   Now, take the square root of this value: $\sqrt{0.36} = 0.6$

   Finally, calculate the de Broglie wavelength λ: $\lambda = \frac{h/e}{\sqrt{\frac{240 m_p k}{r_{min}}}} = \frac{4.2 \times 10^{-15}}{0.6}$ $\lambda = 7 \times 10^{-15} \text{ m}$

5. State the Final Answer in Required Units

   The question asks for the answer in units of femtometers (fm). Since $1 fm = 10^{−15}$ m: $\lambda = 7 \text{ fm}$

   The integer value is 7.