The following three questions from nuclear physics are simple but are useful in your preparation for entrance tests:

(1) Two radioactive samples S

_{1}and S_{2}have half lives 3 hours and 7 hours respectively. If they have the same activity at a certain instant*t*, what is the ratio of the number of atoms of S_{1}to the number of atoms of S_{2}at the instant*t*?(a) 9 : 49

(b) 49 : 9

(c) 3 : 7

(d) 7 : 3

(e) 1 : 1

If The number of atoms present at the instant

*t*is*N*, we have*N = N*

_{0}e

^{–}

*where*

^{λt}*N*

_{0 }is the initial number, e is the base of natural logarithms and

*λ*is the decay constant.

Therefore, activity, d

*N/*d*t*= –*λ**N*_{0}e^{–}*=*^{λt}*λN*If

*N*_{1}and*N*_{2}are the number of atoms of S_{1}and S_{2}respectively when the activities are the same, we have*λ*

_{1}

*N*

_{1}=

*λ*

_{2}

*N*

_{2}from which

*N*

_{1}/

*N*

_{2}=

*λ*

_{2}/

*λ*

_{1}

But the decay constant

*λ*is related to the half life*T*as*T =*0.693/*λ*.Therefore,

*N*_{1}/*N*_{2}=*λ*_{2}/*λ*_{1}=*T*_{1}/*T*_{2}= 3/7 [Option (c)].(2) A nucleus

_{Z}*X*^{A }has mass*M*kg. If*M*_{p}and*M*_{n }denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is(a) [Z

*M*_{p}+ (A – Z)*M*_{n}–*M*]c^{2}(b) [Z

*M*_{p}+ Z*M*_{n}–*M*]c^{2}(c)

*M*– Z*M*_{p}– (A – Z)*M*_{n}(d) [

*M*– Z*M*_{p}– (A – Z)*M*_{n}]c^{2}(e) [A

*M*_{n }–*M*]c^{2}^{}Total mass of the Z protons is Z

*M*_{p}. Since the total number of nucleons is A, the total number of neutrons is (A – Z) and the total mass of the neutrons is (A – Z)*M*_{n}. The mass defect ∆

*M*is the difference between the total mass of the nucleons (protons and neutrons together) and the mass of the nucleus: ∆*M =*[Z*M*_{p}+ (A – Z)*M*_{n}–*M*].Therefore, binding energy.= ∆

*M*c^{2}where ‘c’ is the speed of light in free space.Thus binding energy = [Z

*M*_{p}+ (A – Z)*M*_{n}–*M*]c^{2}(3) If the aluminium nucleus

_{13}Al^{27}has nuclear radius of about 3.6 fm, then the tellurium nucleus_{52}Te^{125}will have radius approximately equal to(a) 3.6 fm

(b) 16.7 fm

(c) 8.9 fm

(d) 6.0 fm.

(e) 4.6

The nuclear radius

*R*is given by*R = R*

_{0}

*A*

^{1/3}where

*R*

_{0}is a constant (equal to 1.2×10

^{–15}m, nearly) and

*A*is the mass number of the nucleus.

If

*R*_{l}and*R*_{2 }are the radii of the given Al and Te nuclei respectively, we have*R*

_{l}=

*R*

_{0 }(27)

^{1/3}= 3

*R*

_{0}and

*R*

_{2}=

*R*

_{0 }(125)

^{1/3}= 5

*R*

_{0}

Dividing,

*R*_{l}/*R*_{2}= 3/5Therefore,

*R*_{2}= 5*R*_{1}/3 = (5×3.6)/3 fm = 6 fm.By clicking on the labe l ‘nuclear physics’ below this post, you can access all posts related to nuclear physics on this site.

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