Question 1 One form of nuclear radiation, beta decay, occurs when a neutron chan

Question

Question 1 One form of nuclear radiation, beta decay, occurs when a neutron changes into a proton, an electron and a neutral particle called a neutrino. When this change happens to a neutron within the nucleus of an atom, the proton remains behind in the nucleus while the electron and neutrino are ejected from the nucleus. The ejected electron is called a beta particle. One nucleus that exhibits beta decay is the isotope of hydrogen 3H, called tritium, w hose nucleus consists of one proton making it hydrogen and two neutrons (giving tritium an atomic mass m-3 u Tritium is radioactive, and it decays to helium. Suppose an electron is ejected from a 3H atom, which has a radius of 1.000x10-14 m. The resulting 3He atom has the same radius as the 3H atom. What is the escape velocity of the electron ejected from the process? Note: Your answer may be larger than the speed of light which is okay in this scenario. To solve this problem correctly we would need to use special relativity

Explanation / Answer

given
an electron escapes from 3H atom
radius of 3H atom, r1 = 10^-14 m
rdius of 3He atom is the same
let the escape spee dof the electron be u

now, from conservation of energy
energy produced in the reaction = mass deficit * speed of light squared

now, mass deficit = -(mass of products - mass of reactants)
mass of products = mass of 3He + mas sof electron + mas sof neutrino
mass of 3He = 3.02u
mass of electron = 0.00054858u
mass of neutriono = 0 ( massless)
mass of reactants = mass of 3H = 3.016049u
hence
mass deficit = 0.00449958u
hence
energy produced, E = (0.00449958*1.6*10^(-27))(3*10^8)^2 = 6479395.2*10^-19 J

now, from conservation of momentum we can say that the 3He product will have negligible speed when compared to the electron as electonn has very small mass
hence
from conservaiton of energy
0.5*9.1*10^(-31)*u^2 = 6479395.2*10^-19 J
u = 11.93332*10^8 m/s

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