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**SHORT ANSWER TYPE QUESTIONS**

**Q.16. Why do stable nuclei never have more protons than neutrons?Ans.** The reason is that protons, being charged particles, repel each other. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.

To view this in quantum mechanical terms, the proton potential well is not as deep as the neutron well due to the electrostatic repulsion. [Due to the Pauli exclusion principle, you only get two particles per level (spin up and spin down)]. If one well is filled higher than the other, you tend to get a beta decay to even them out. As the nuclei get larger, the neutron well gels deeper as compared to the proton well and you get more neutrons than protons.

**Q.17. Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence:A â†’ B â†’ CHere B is an intermediate nuclei which is also radioactive.Considering that there are N**Consider radioactive nucleus A have N

Ans.

T

According to radioactive decay law,

N=N

12=16e

**Q.19. Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10 ^{â€“15} m.Ans. **To detect the properties of nucleons inside the nucleus the wavelength of particle which may detect nucleons that must be of size of nucleons (10

Î» = 10

Î» = h/p

âˆµ E = hv = hc/Î» [âˆµc = vÎ»]

=12.4Ã—10

K.E. = 1.24Ã—10

So the K.E. of particle which may detect nucleon inside the nucleus must be of 1.24Ã—10

(a) What nuclide is a mirror isobar of

(b) Which nuclide out of the two mirror isobars have greater binding energy and why?

Ans.

Here Z is atomic number and N is no. of neutron in

Z

N

Mirror isobar of

Z

So Mg is isobar of

So

(b) As the neutrons in

So,

**LONG ANSWER TYPE QUESTIONS**

**Q.21. Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is:Assume that we start with 1000 ^{38}S nuclei at time t = 0. The number of ^{38}Cl is of count zero at t = 0 and will again be zero at t =âˆž . At what value of t, would the number of counts be a maximum?Ans.**

Initially at t=0, number of radioactive atoms of S

At any time t,

and N

It is the rate of formation of Cl

=Î»

Multiplying both sides by e

Integrating both sides

âˆ´ Cl

at t=0, N

Multiplying E

N

No. Are Cl

Put the value of N

By cross multiplication and multiplying both sides by

Number Cl

Ans.

Some part of energy of Î³-ray is used up against binding energy B = 2.2 MeV and the rest part will impart K.E. to neutron and proton.

By law of conservation of momentum,

âˆ´ P

It can be possible if p

If p

It is a quadratic equation so its solution by quadratic formula

For a real and equal value of P

âˆµ Î» is very small

So E=B

If proton and neutron had charge eâ€™ each and governed by the same electrostatic force, then in the above equation we would need to replace electronic mass m by the reduced mass mâ€™ of proton-neutron (as some mass of proton and neutron is used by binding energy) and electronic charge e is replaced by eâ€™.

=M/2 ( if m = mass of electron)

m' = 1836m/2 = 918 m

âˆ´ Binding Energy

Dividing (II) by (i) we get,

e'/e = 2200000/1248.48 = (176.21)

Required ratio e'/e = 3.64

If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them.Experimentally, the electron energy was found to have a large range.

Ans.

By the law of conservation of momentum,

P

O= p

Let p

â‡’ |p

Energy of proton = E

Energy of electron

From conservation,

âˆ´ By the law of conservation of energy,

and m

and m

As the energy difference in neutron and proton is very small, pc will be small pc<<m

Again pc<<m

Pc = m

Pc=2 MeV is the momentum

E=mc

âˆµ E

E is the energy of either proton or neutron then

**(i) Plot the graph of R versus t and calculate half-life from the graph.(ii) Plot the graph of versus t and obtain the value of half-life from the graph.Ans.**

(i) Graph between R versus t is exponential curve. From the graph at slightly more the R should be 50% so at R =50% the t(h)=0.7h

=0.7Ã—60 min

=42 min

(ii) For Graph between versus t(h)

=2.302 log

at t=2 hours,

= 2.303log

t (hours) | 1 | 2 | 3 | 4 |

-1.04 | -2.08 | -311 | -4.16 |

The graph showing the variation of versus t(h) as follows:

We know that disintegration constant

Î» = -1.05 per hour

t_{1}_{/2} = 0.6931/Î» = 0.6931/1.05

t_{1}_{/2 }= 42 min**Q.26. Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable.(i) Verify this by calculating the proton separation energy Sp for ^{120}Sn (Z = 50) and ^{121}Sb = (Z = 51).The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given byS_{p} = (M_{Zâ€“1, N} + M_{H} â€“ M_{Z,N} )c^{2}.Given ^{119}In = 118.9058u, ^{120}Sn = 119.902199u, ^{121}Sb = 120.903824u, ^{1}H = 1.0078252u.(ii) What does the existence of magic number indicate?Ans.**

(i) S

Here in this formula M

M

âˆ´ M

It is

S

S

Now for S

S

Z=51, Z-1=50 for S

M

âˆ´ S

= 0.0059912c

âˆ´ S

(ii) The existence of magic numbers indicates that the shell structure of nucleus is similar to the shell structure of atom. This also explains the peaks in binding energy per nucleon curve.

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