InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The most abundant isotope of helium has a `._(2)^(4)H` nucleus whose mass is `6.6447 xx 10^(-27) kg` . For this nucleus, find (a) the mass defect and (b) the binding energy. Given: Mass of the electron: `m_e =5.485799 xx 10^(-4) u`, mass of the proton: `m_(P) =1.007276 u` and mass of the neutron: `m_(n) =1.008 665 u`. |
|
Answer» The symbol `._2^4He` indicates that the helium nucleus contains `Z = 2` protons and `N = 4-2=2` neutrons. To obtain the mass defect `Delta m`, we first determine the sum of the individual masses of the separated protons and neutrons. Then, we subtract from this sum the mass of the mass of the nucleus. (a) We find that the sum of the individual masses of the nucleons is `underbrace (2(1.6726 xx 10^(-27) kg))_("two protons") + underbrace (2(1.6749 xx 10^(-27) kg))_("two neutrons")` `=6.6950 xx 10^(-27) kg` This value is greater than the mass of the intact He necleus, the mass defect is `Delta m=6.6950 xx 10^(-27) kg -6.6447 xx 10^(-27) kg` `=0.0503 xx 10^(-27) kg` (b) According to Eq. (iv), the binding energy is given by `Delta E_(BE) =(Delta m)c^(2) =(0.0503 xx 10^(-27) kg) (3.00 xx 10^(8) ms^(-1))^(2)` ` =4.53 xx 10^(-12) J` Usually, binding energies are expressed in energy units of electrons volt instead of joule `(1 eV =1.60 xx 10^(-19) J)`. because Binding energy `=(4.53 xx 10^(-12) J)((l eV)/(1.60 xx 10^(-19) J ))` `=2.83 xx 10^(7) eV =28.3 MeV` In this result, one million electron volt is denoted by the unit MeV. The value of `28.3 MeV` is more than two million times greater than the energy required to remove an orbital electron from an atom. |
|