A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.
The speed of daughter nuclei is
B.
Conserving the momentum
Now, from energy conservation and mass -energy equivalence
A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. The speed of light is c.
The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then
E1 = 2E2
E2 = 2E1
E1 > E2
E1 > E2
D.
E1 > E2
After decay, the daughter nuclei will be more stable hence, binding energy per nucleon will be more than that of their parent nucleus.
The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions:
(i) A + B → C + ε (ii) C → A + B + ε
(iii) D + E → F + ε and (iv) F → D + E +ε
where ε is the energy released? In which reaction is ε positive?
(i) and (iv)
(i) and (iii)
(ii) and (iv)
(ii) and (iv)
A.
(i) and (iv)
Binding energy per nucleon of each product is less than that of each reactant.
A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α–particles and 2 positions. The ratio of number of neutrons to that of protons in the final nucleus will be
B.
In positive beta decay a, proton is transformed into a neutron and a positron is emitted
p+ → n0 + e+
Number of neutrons initially was A-Z
Number of neutrons after decay (A-Z) -3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay)
The number of protons will reduce by 8. so, the ratio number of neutrons to that of protons =