Class 12-science H C VERMA Solutions Physics Chapter 24 - The Nucleus
The Nucleus Exercise 442
Solution 1
We know,
Also,
In CGS system,
Specific gravity=
Solution 2
We know,
R=15km
Solution 3
And Binding energy = 
Binding energy=28.2MeV
Solution 4
As given:
7
Energy,
E=7Li
=0.018076u
E=16.83MeV
Solution 5
Binding energy is:
For Au binding energy per nucleon
=7.73
Solution 6
(a) 238U→2He4+Th234
energy is:
E=[Mu-(NHe+ MTh)]u
=[238.0508-(234.04363+4.00260)]u
E=4.255MeV
(b) Energy is:
E=23.007MeV
Solution 7
223Ra→209Pb+14C
=223.018-(208.981+14.003)
=0.034
Energy is:
E=31.65MeV
Solution 8
Hydrogen contains only protons. Thus reaction will be:
∴ Minimum energy required is:
Solution 9
Energy released = (Initial mass of nucleus-Final mass of nucleus)C2
=(MZ.N-1+M0-MZN)C2
Solution 10
Energy is :
E=(31.974-31.972)u
=1.86MeV
Solution 11
(a) We know,
(b) Energy is:
E=[Mn-(MP+Me)]u
=[Mnu-Mpu-Meu]C2
=782159eV
=782.1keV
Solution 12
Solution 13
(a) EN=EP-K.E.
= 0.650-0.150=0.5MeV
= 500keV
(b) Momentum 
Solution 14
(b) We know
Q=[Mass of reactants - mass of products]C2
Now for equation 1
Q=[39.9640u-39.9626u]C2
=1.3034MeV
For equation 2
Q=[39.9640u-39.9624u]C2
=0.4676MeV
For equation 3
Q=[39.9640u-39.9624u]uC2
=1.490MeV
Solution 15
Solution 16
Energy liberated =Mc-MB
=11.0114u-11.0093u
=933.6keV
Solution 17
Mass of 234Ra 
=208563.0195MeV
=5.304MeV
Solution 18
According to question the reaction will be :
12N → 12C+e++v + v
Thus , energy is given as:
E(e++v)=E(N)-E(C12+v)
=12.89MeV
Solution 19
We know,
(a)
=64min
(b) Average life
=92min
c) 
T=1598.23s
The Nucleus Exercise 443
Solution 20
(a) 198g of Au=Nr atoms
Activity =
= 0.244curies
(b) We know,
= 0.040curies
Solution 21
(a) We know,
(b)
Solution 22
(a) 
(b) Half-life of uranium, 
(c) We know,
Solution 23
(a) We know,
(b) 
Solution 24
Also,
Solution 25
1 Mole of RaCl2=226+71=297g
Now
And activity=
Solution 26
After 9 hours
=0.536Ci
Now, no. of atoms after 9th hour
Activity after 10 hours
=0.5Ci
Atoms left after 10th hour
Total no. of disintegration in tenth hour is :
Solution 27
We know,
=186.9
Solution 28
Emission rate is reduced to half
from 
∴ the
amount of emission of cobalt should be reduced to half.
Thus, the required time is 270days.
Solution 29
(a) 
Thus, it is 𝛽+ decay.
(b) Initially for carbon 90 N0 and boron 10 N0 and finally for carbon 10 N0 and boron 90 N0.
Solving we get
t=64.3
Solution 30
(a) We know,
Activity 
(b) Number of decays in 10 hrs is:
(c) We know,
Now,
No. of atoms disintegrated is:
Solution 31
Total counts radiated = counts received per cm2× Total area
Also,
Solution 32
Initially,
Total no. of uranium atoms= No. of atoms of U + No. of atoms of Pb
Also
Solution 33
We know,
After 't' time is passed,
Solution 34
According to question :
Activity of bottle on mountain = 1.5% × Activity of bottle 8 years before
t= 83.75 years
Solution 35
(a) 
b) From graph,
c) Half-life,
Solution 36
We know,
A=dN
Now if
And 
Also, abundance in potassium 
Solution 37
(a) Here,
Neutrino is emitted is the decay.
(b) We know,
Solving we get,
Solution 38
Rate of radioactive decay is given as:
According to the question, after some time the number of active nuclei will become constant. Thus,
Also,
Solution 39
Number of radioactive particles present at 't' time is given as:
And particles that decay are given as:
Solution 40
We know,
Also
Now,
Fraction of activity 
Solution 41
No. of atoms is:
Activity 
Or 729Ci
The Nucleus Exercise 444
Solution 42
(a) 
b) At t=/hr
Bi will be decayed to half as 
Now,
Mass of 
=0.175g
Similarly for 
Mass of 
=0.325g
Solution 43
(a) We know, 
disintegration/s
Now we will find 
for all values.
(b) 
for 100 Ag = 24.4s
(c) 
=0.0284
And also,
(d)
(e)According to above graph 
108Ag)
Solution 44
We know,
= 4.8h
Also 
Solution 45
We know
Solution 46
After 't' time
Solution 47
1g of I =0.007g U235
And
1 atom=200MeV
Solution 48
Power = Total energy emitted per seem
And we know that
=3.7mg
Solution 49
(a) No. of fission per second=
(b) 
and so 
=1.26kg/day
Solution 50
(a) Q value 
=4.05MeV
(b) Q value 
=3.25MeV
(c) Q value 
=17.57MeV
Solution 51
We know,
And 
Solution 52
Q value = 
=-93.1KeV
Solution 53
% of deuterium
Energy of deuterium 
=[2
(2H)-m(3H)-mp]C2
=[22.014102-3.016049-1.007276]uC2
=4.54MeV or 7.262
