A) 731 MeV
B) 7.31 MeV
C) 0.731 MeV
D) 1.17 MeV
Correct Answer: C
Solution :
From Einsteins mass energy relation \[\Delta E=(\Delta m){{c}^{2}}\] Mass of proton \[=1.6725\times {{10}^{-27}}kg\] Mass of electron \[=\underline{0.0009\times {{10}^{-27}}}kg\] Their sum \[=\underline{1.6734\times {{10}^{-27}}}kg\] Mass of neutron \[=1.6747\times {{10}^{-27}}kg\] Their difference \[\Delta m=0.0013\times {{10}^{-27}}kg\] \[\therefore \] \[\Delta E=(0.0013\times {{10}^{-27}}){{(3\times {{10}^{8}})}^{2}}\] \[=1.17\times {{10}^{-13}}J\] Also, \[1.6\times {{10}^{-19}}J=1eV\] \[\therefore \] \[\Delta E=\frac{1.17\times {{10}^{-13}}}{1.6\times {{10}^{-19}}}=0.731\times {{10}^{6}}eV\] \[=0.731MeV.\]You need to login to perform this action.
You will be redirected in
3 sec