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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=

We know,

R=15km

### Solution 3

And Binding energy =

Binding energy=28.2MeV

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) 238U2He4+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

223Ra209Pb+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

(a) We know,

(b) Energy is:

E=[Mn-(MP+Me)]u

=[Mnu-Mpu-Meu]C2

=782159eV

=782.1keV

### 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 16

Energy liberated =Mc-MB

=11.0114u-11.0093u

=933.6keV

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

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

(a) We know,

(b)

### Solution 22

(a)

(b) Half-life of uranium,

(c) We know,

(a) We know,

(b)

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 :

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:

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

(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

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)

We know,

= 4.8h

Also

We know

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

(a) Q value

=4.05MeV

(b) Q value

=3.25MeV

(c) Q value

=17.57MeV

We know,

And

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