Question

1.3 In the period before the discovery of the neutron, many people thought that the nucleus onsisted of protons and electrons, with the atomic number equal to the excess number of protons. Beta decay seemed to support this idea - after all, electrons come popping out; doesn't that imply that there were electrons inside? Use the position-momentum uncertainty relation, Ax Ap 2 h/2, to estimate the minimum momentum of an electron confined to a nucleus (radius 10-13 cm). From the relativistic energy-momentum relation. E2-pe-m2c4, determine the corresponding energy and compare it with that of an electron emitted in, say, the beta decay of tritium (Figure 1.5). (This result convinced some people that the beta decay electron could not have been rattling around inside the nucleus, but must be produced in the disintegration itself.)

Answer 1

We know that Delta x Delta_p = hbar/2 Delta_p = hbar/2 Delta x where Delta x = 10^-15 m The size of atomic nucleus Delta p = hbar/2 Delta x Delta_p = 1.05 times 10^-34/2 (10^-15) Delta_p = 5.25 times 10^-20 kg m s^-1 e^- inside Nucleus can have minimum momentum P = 5.25 times 10^-20 kg ms^-1 Now ^2 = P^2 c^2 + m_0^2 C^4 E^2 = (5.25 times 10^-20)^2 (3 times 10^8)^2 + m_0^2 c^4 where m_0 c^2 Vertical-bar _e^- = 511 kev = 5.11 times 10^5 ev = 8.176 times 10^-14 J Because ev = 1.6 times 10^-19 J E^2 = (5.25 times 10^-20)^2 (3 times 10^8)^2 + (8.176 times 10^-14)^2 E^2 = 2.48 times 10^-22 E = squareroot 2.48 times 10^-22 E = 1.57 times 10^-11 J \r\n E = 9.84 times 10^7 ev E = 98.4 mev Average energy of e^- in Beta decay of is 5.7 kev < < = 98.4 mev Hence e^- cannot find inside Nucleus