Question

Suppose an attractive nuclear force acts between two protons which may be written as F=Ce−kr/r2. (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 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.

Answer 1

Answer:

between two protons which may be written as F=Ce−kr/r2. (a

Answer 2

Explanation:

( a) Given in the question

Step 1:

$F=\frac{C e^{-k \tau}}{r^{2}}$

Where  

F is nuclear force of attraction,  

Step 2:

Here e−Kr is just a pure number, that is a quantity without dimensions.

$[c]=[F] \times\left[r^{2}\right]$

$[C]=\left[\mathrm{MLT}^{2}\right] \times\left[\mathrm{L}^{2}\right]$

$[C]=\left[M L^{3} T^{-2}\right]$

And $[K]=\frac{1}{[r]}=\left[L^{-1}\right]$

(b) Electric force by Coulomb's Law,    

Step 1:      

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$  

Take $r=5 \times 10^{-15} \mathrm{m}$

Then we get

$=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5 \times 10^{-10}\right)^{2}}$  

$=9 \times 10^{9} \times \frac{2.56 \times 10^{-38}}{25 \times 10^{-20}}$

$=9 \times 10^{9} \times 0.1024 \times 10^{-18}$

$=0.9216 \times 10^{-9}$

$=9.216 \times 10^{-10}$

Step 2:

nuclear force is $(F)=\frac{c_{e}^{-k r}}{r^{2}}$  

Take $r=5 \times 10^{-15}$m and$k=1 \text { fermi }^{-1}$

$F=\frac{C \times 10^{-5}}{\left(5 \times 10^{-10}\right)^{2}}$

Step 3:  

When we equate the two forces we get

$9.216 \times 10^{-10}=\frac{C \times 10^{-5}}{\left(5 \times 10^{-10}\right)^{2}}$

$9.216 \times 10^{-10}=\frac{C \times 10^{-5}}{25 \times 10^{-20}}$

$9.216 \times 10^{-10} \times 25 \times 10^{-20}=C \times 10^{-5}$

$\frac{9.216 \times 10^{-10} \times 25 \times 10^{-20}}{10^{-5}}=c$

$C=2.304 \times 10^{-23} \mathrm{N}-\mathrm{m}^{2}$