NCERT Solution Physics Class 12 Chapter 13 Nuclei
NCERT Solution Physics Class 12 Chapter 13 Nuclei all questions and answers. Physics Class 12 13th Chapter Nuclei exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 13. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 13 Nuclei.
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ANSWER-
We know that in every α-decay there is loss of 2 proton and 4 neutron. In β+decay there is a loss of 1 proton and a neutrino (n).In β– decay there is a gain of 1 proton and an antineutrino (v̂) is emitted from the nucleus.
Nuclear reactions for all the scenarios mentioned below
(i) We know that in every α-decay there is loss of 2 proton and 4 neutron. hence
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13.21 From the relation R = R0 A 1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
ANSWER-
Given relation is
R = R0 A 1/3, where R0 is a constant and A is the mass number of a nucleus
Nuclear matter density is given by,
p = (mass of the nucleus)/(volume of the nucleus)
∴ p = M/V
Where, mass of the nucleus (M) = average mass of nucleus(m) x mass number of a nucleus (A)
Volume of the nucleus (V) = 4/3 πR3 .
From above formula we can conclude that nuclear matter density is independent of mass number.
ANSWER-
During electron capture process there is release of energy E1 then
Let me is mass of electron.
Using mass defect, energy releases during electron capture E1 is given by,
E1 = △m x c2
Where △m is mass defect in the process = mN1 +me – mN2 .
∴ E1 = [mN1+me – mN2 ] x c2
But for electron capture
Nuclear mass of nucleus () = atomic mass – mass of all electrons
∴ mN1= m1 – Zme and mN2 = m2-(Z-1)me where Z is number of electrons.
∴ E1 = [m1 – Zme +me -m2+ (Z-1)me] x c2
∴ E1=[m1-m2] x c2…..(1)
For the β + (positron) emission from a nucleus energy release is E2 is given by
∴ E2= [mN1– me – mN2 ] x c2
∴ E2 =[m1 -m2-2me] x c2……(2)
From equations 1 and 2
If E2>0 then from equation 2
∴ [m1 -m2 -2me ]>0
∴ [m1 -m2]>0
Therefore from equation 1 we can write E1>0 .
But if E1>0 then from equation1,
∴ [m1 -m2] > 0
But we can’t write from this that E2>0. Hence from this we can conclude that if β + emission is energetically allowed (E2>0 ), electron capture is necessarily allowed but not vice–versa.
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