PROBLEM ON NUCLEAR PHYSICS
Q/1. The measured
mass of deuteron atom(2
1H) , Hydrogen atom(1
1H) , proton and neutron
is 2.01649 u ,1.00782 u , 1.00727 u and 1.00866 u . Find the binding energy of
the deuteron nucleus (unit MeV /nucleon).
Ans: Here A=2, Z=1, N=1
B E = [ZMH + NMN + M(11H)]✖ 931.5 MeV
= [1✖ 1.00782 +1✖ 1.00866 -2.01649] ✖931.5 MeV
= [0.00238]✖ 931.5 MeV
=2.224 MeV
Q/2. The binding energy of the neon isotope (2010Ne)
is 160.647 MeV. Find its atomic mass?
Ans: Here A=10, Z=10,
N=10
M (A
ZX) = [ZMH + NMN] – B/(931.5 MeV/u)
M (A ZNe) =[10 ✖1.00782 + 10 ✖ 1.00866] -160.647 /931.5
MeV/u
=
19.992u
Q/3. (a) Find the
energy needed to remove a neutron from the nucleus of the calcium isotope (4220Ca) .
(b) Find the energy needed to remove a proton from this
nucleus.
(c) Why are these
energies different?
Given: Atomic masses of 4220Ca
= 41.958622u, 4120Ca= 40.962278u, 41 19K
= 40.961825u, and mass of 10n
=1.008665u, 11P
=1.007276u .
Ans: (a) 4220Ca ➡ 4120Ca
+10n
;
Total mass of the 4120Ca & 10n
= 41.970943u
Mass defect Dm=41.970943 -41.958622=
0.012321u
So, B.E. of missing neutron = Dm ✖931.5MeV = 11.48MeV
(b) 4220Ca ➡ 41 19K
+11P;
Total mass of the 41 19K & 11P41.919101u
Mass defect Dm= 41.919101 - 41.958622= 0.010479u
So, B.E. of
missing protron = D
m✖ 931.5MeV = 10.27MeV
(c) The neutron was
acted upon only by attractive nuclear forces whereas the proton was also acted
upon by repulsive electric forces that decrease its binding energy.
Q/4. Half life of P is 14.3 days. If you have 1.00
g of P today, then what would be the amount remaining in 10 days?
Ans: t1/2
=0.693/
λ= 0.693/14.3= 0.04847 per day
N= N0e-λt ⇒N = (1)e-0.04847*10
⇒N = 0.616g
Q/5. A radioactive
nucleus has a half life of 100 years. If the number of nuclei t = 0 is N0
, then find the
number of nuclei that have decayed in 300 years?
Ans : Number
of nuclei present after 300 year N=N0(1/2)(T/T1/2)
N' = N0 - N0 (1/2)3
= (7/8) N0
Q/6. The atomic
ratio between the uranium isotopes 238U
and 234U in a mineral sample is found to be 1.8 ´ 104 . The half
life of 238U is 4.5 ´109 years, then find
the half life of 234U .
Ans: NAλA=NB λB
Þ NA/NB =λA/λB =T1/2A/T1/2B
Þ T1/2B = (NA/NB ) T1/2A =4.5´109 /1.8 ´ 104 = 2.5 ´ 105
Q/7: A radioactive sample contains 1.00 mg of radon 222Rn,
whose atomic mass is 222 u . The half life of the radon is3.8 day . Then find
the activity of the radon?
Ans: Decay constant λ = 0.693/ T1/2 =
0.693/(3.8✖24✖60✖60)=2.1✖105years
Number of atoms in 1.00 mg is N= 1✖10-6kg/{(222u)✖1.66✖10-27kg/u}
N= [ 1✖10-6kg/{(222u)✖1.66✖10-27kg/u}]✖6.023✖1023 = 2.7✖1018atoms
Hence, activity R= Nλ =2.1✖10-6 ✖2.7✖1018 =5.7✖1012 decay/sec
Thank you sir ❤️
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