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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

=[2 2.014102-3.016049-1.007276]uC2

=4.54MeV or 7.262  

 

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