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Artificial radioactivity and application of radioactivity

Q: After \$280\$ days, the activity of a radioactive sample is \$6000 dps\$. The activity reduces to \$3000 dps\$ after another \$140 days\$. The initial activity of the sample in \$dps\$ is:

Here, the activity of the radioactive sample reduces to half in \$140\$ days. Therefore, the half life of the sample is \$140\$ days. \$280\$ days is it's two half lives. So before two half lives it's activity was\${2^6} \times 6000 = 24000\,dps.\$Hence, the option \$D\$ is the correct answer.

Q: What Is the longest wavelength of a photon that can ionize a hydrogen atom in its ground state? Specify the type of radiation.

Since the energy of the incident photon\$=hv=\dfrac{hc}{\lambda}=13.6\,eV\$\$\lambda =\dfrac{6.6\times10^{-34}\times3\times10^8}{13.6\times 1.6\times10^{-19}}\$\$\lambda=0.910 \times 10^{-10}\,m\$This radiation is in the ultraviolet region.

Q: If the activity of \$^{108}\$\$Ag\$ is \$3\$ micro curie, the number of atoms present in it are ( \$\lambda =\$ 0.005 sec\$^{-1}\$)

\$\lambda = 0.005 sec^{-1}\$\$ 1\$ Micro curie \$= 3.7\times10^{4} dps\$Assuming all \$Ag\$ atoms to be radioactive, and each showing Let no of atoms present \$ = 1 \ dps \ no \$Given:\$(\dfrac{dN}{dt}) activity= \lambda N_0 \Rightarrow 3.7\times 3\times 10^{4} dps = 0.005\times N_0 \$ \$ \Rightarrow N_0 = 2.2\times 10^{7} atoms\$

Q: The mass of U\$^{234}\$ of radioactivity \$1\$ curie is

1 Curie \$ = 3.7\times10^{10} dps\$Assuming, atom of \$U^{234}\$ shows radioactivity and is at its radioactive state. Mass of 1 amu \$ = 1.66\times10^{-27} kgs\$Mass of 1 curie \$ = 3.7\times10^{10} \times234\times1.66\times10^{-27}\times10^{3} gms\$ \$ = 1.4372 \times10^{-11} gms \approx 1.438\times10^{-11} gms.\$

Q: The increasing order of specific charge for electron(\$e\$), proton(\$p\$), neutron(\$n\$) and alpha particle(\$\alpha \$) is

Specific Charge = \$\dfrac{q}{m}\$Charge on neutron = 0\$SC_{neutron} \$ = 0\$SC_{proton} = \dfrac{e}{m_P} \$\$SC_{\alpha - particle} = \dfrac{2e}{4m_P} \$\$SC_{electron} = \dfrac{e}{10^{-3} m_P} \quad [m_e \approx 10^{-3} m_P]\$ Hence, the specific charges in increasing order are \$ n , \alpha, p, e \$

Q: 'Radioactivity is a nuclear phenomenon'. Comment on this statement.

Any physical change (such as change in pressure and temperature) or chemical change (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) does not change the rate of decay and the nature of radiation emitted by the substance. This shows clearly that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. Therefore the radioactivity should be the property of the nucleus. Thus the radioactivity is a nuclear phenomenon.

Q: One curie is equal to

By definition, one curie is defined as \$3.7 \times 10^{10}\$ atoms decay in one second. Curie is a non SI unit of radioactivity.\$\therefore 1 Ci = 3.7 \times 10^{10}\$ dps

Q: What is radioactivity?

The act of emitting radiation spontaneously, as its name implies, is known as radioactivity. An atomic nucleus that is unstable for some reason "wants" to give up some energy in order to shift to a more stable configuration achieves this.Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting ionizing radiation. As each nucleus disintegrates, it generates a charged particle with enough kinetic energy to penetrate solid matter.

Q: Differentiate between half life and average life of a radioactive substance.

The difference between half life and average life is given below :- Half lifeAverage life (i) The time required for half of the radioactive substance to decay is called as half life of that substance.(ii) It is denoted by \$t _{\frac{1}{2}}\$(iii) \$t_{\frac{1}{2}} = \$ half life \$= \left(\dfrac{\ln 2}{\lambda} \right)\$where \$\lambda\$ is decay constant.(i) The expected life for the radioactive substance within a sample is called as average life of that substance.(ii) It is denoted by \$\tau\$.(iii) \$\tau =\$ average life \$= \dfrac{1}{\lambda}\$

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