Question

The atomic mass of F19 is 18.8884 u. If the masses of proton and neutron are 1.0078

u and 1.0087 u respectively, calculate the binding energy per nucleon.​

Answer 1

The binding energy=0.4017×$10^{-11}$J

Step-by-step explanation:

No of proton of fluorine is 9

No of neutron of  fluorine is 10

$M_{d}=(m_{n}+m_{p} ) - m_{o}$

     =((10×1.0087)+(9×1.0078)) - 18.8884

      =0.2688 amu

The mass of fluorine = 0.2688×$\frac{1}{6.02214*10^{26} }$ kg

                                  =0.4463×$10^{-27}$kg

The binding energy E = m$c^{2}$

                                      =0.4463×$10^{-27}$×(3×$10^{8}$)^2  J

                                     =0.4017×$10^{-11}$J