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प्रश्न
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उत्तर
The substance's rate of decay and the type of radiation it emits are unaffected by any physical or chemical changes, including variations in temperature and pressure, extreme heating, freezing, the effect of strong electric and magnetic fields, chemical treatment, oxidation, etc.
This demonstrates unequivocally that the orbital electrons, which are susceptible to such alterations, cannot be the cause of the radioactive event. Thus, the nucleus should be the source of radioactivity. The phenomenon of radioactivity is thus nuclear.
संबंधित प्रश्न
What happens to the atomic number of element when (i) An α -particle, (ii) A β -particle 1 and (iii) γ-radiation is emitted?
Draw a diagram of a simple atom showing the nucleus and electrons. In this atom:
(a) What type of charge will there be on the nucleus?
(b) What is the value of this charge?
What is nucleus of an atom? Compare its size with that of the atom. Name its constitutents. How is the number of these constituents determined by the atomic number its atomic model.
Are all the radiations viz. α, β and γ emitted in a single radioactive decay?
Which of the following statements is/are correct?
- α particles are photons
- Penetrating power of γ radiation is very low
- Ionization power is maximum for α rays
- Penetrating power of γ radiation is very high
A radioactive nucleus \[\ce{^A_ZX}\] undergoes spontaneous decay in the sequence \[\ce{^A_ZX -> _{Z-1 }B -> _{Z-3}C -> _{Z-2}D }\], where Z is the atomic number of element X. The possible decay particles in the sequence are ______.
Two radioactive sources A and B of half lives of 1 hour and 2 hours, respectively, initially contain the same number of radioactive atoms. At the end of two hours, their rates of disintegration are in the ratio of ______.
The mean lives of a radioactive substance are 1620 and 405 years for β-emission and β-emission respectively. The time after which three fourth of a sample will decay if it is decaying both by β-emission and β-emission simultaneously will be ______ years.
(Take ln 2 = 0.693)
Complete the following radioactive reaction:
\[\ce{^-_-X -> ^-_-Y + ^4_2He -> ^234_91Z + ^0_-1e}\]
