HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q15.2 Solution: Suppose an attractive nuclear force acts between two protons…

QPages 121-124: Q15.2: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q15.2 Solution: Suppose an attractive nuclear force acts between two protons…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the force law is written as $F=C,e^{-kr}/r^2$, the product $kr$ must be dimensionless. Students often plug in $k=1; ext{fm}^{-1}$ and $r=5; ext{fm}$ directly, treating $k$ as $1; ext{m}^{-1}$, which makes $e^{-kr}approx e^{-5 imes10^{-15}}approx1$ and yields an enormous $C$. Convert $k$ to SI units first: $k=1; ext{fm}^{-1}=10^{15}; ext{m}^{-1}$, then $kr=5$ and $e^{-kr}=e^{-5}approx0.0067$. Also remember that $C$ carries units of $N,m^2$, not $N,m$.

Suppose an attractive nuclear force acts between two protons which may be written as

F = C e^{-kr}/r^2

. Suppose that

k = 1 \text{ fermi}^{-1}

and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

Step-by-Step Verified Solution

**Answer:**

C = 3.2 \times 10^{-26}

N m_. **Step-BY-STEP Explanation:** At

r = 5

fermi, the nuclear force balances the electric force. Electric force:

F_e = \frac{k e^2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(5 \times 10^{-15})^2} = 9.2

N. Nuclear force:

F_n = \frac{C e^{-kr}}{r^2}

with

k = 1

fermi__ =

10^{15}

m__. At

r = 5

fermi,

e^{-kr} = e^{-5} = 0.00674

9.2 = \frac{C \times 0.00674}{(5 \times 10^{-15})^2} \Rightarrow C = \frac{9.2 \times 25 \times 10^{-30}}{0.00674} = 3.4 \times 10^{-26} \text{ N m}^2

QPages 121-124: Q15.2: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q15.2 Solution: Suppose an attractive nuclear force acts between two protons…

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