_{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*?

*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.

*N/*d

*t*= –

*λ*

*N*

_{0}e

^{–}

*=*

^{λt}*λN*

*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}

*λ*is related to the half life

*T*as

*T =*0.693/

*λ*.

*N*

_{1}/

*N*

_{2}=

*λ*

_{2}/

*λ*

_{1}=

*T*

_{1}/

*T*

_{2}= 3/7 [Option (c)].

_{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

*M*

_{p}+ (A – Z)

*M*

_{n}–

*M*]c

^{2}

*M*

_{p}+ Z

*M*

_{n}–

*M*]c

^{2}

*M*– Z

*M*

_{p}– (A – Z)

*M*

_{n}

*M*– Z

*M*

_{p}– (A – Z)

*M*

_{n}]c

^{2}

*M*

_{n }–

*M*]c

^{2}

^{}

*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}.

*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*].

*M*c

^{2}where ‘c’ is the speed of light in free space.

*M*

_{p}+ (A – Z)

*M*

_{n}–

*M*]c

^{2}

_{13}Al

^{27}has nuclear radius of about 3.6 fm, then the tellurium nucleus

_{52}Te

^{125}will have radius approximately equal to

*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.

*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}

*R*

_{l}/

*R*

_{2}= 3/5

*R*

_{2}= 5

*R*

_{1}/3 = (5×3.6)/3 fm = 6 fm.