Atomic Physics by Rebecca Childs.

By Rebecca Childs.

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Thus an α-particle may have to try for 10 38 = 1017 sec = 3 × 109 years before it actually escapes. 5 billions). 0001 sec. There are two reasons for it: (i) An α-particle in Po214 nucleus has a large energy. (ii) The height of the potential barrier is smaller. Therefore the chance of escape in a single collision is much larger, 1 out of 1017. 0001 sec. 10 21 Explanation of β-emission from Radioactive Nuclei: Like α-decay, the β-decay and positron emission are the means by which a nucleus alters its composition (neutron/proton ratio) to achieve greater stability.

Thus the process of internal conversion, which gives rise to β-ray line spectrum, is a direct transfer of excitation energy from the nucleus to one of the surrounding electrons. Since in the internal conversion the electrons do not come from the nucleus, the not a true form of β-disintegration. 4 A). They have no electric charge, and so they cannot be deflected by magnetic or electric fields. Consequently, direct measurements of their energies (or wavelengths) with a magnetic spectrometer are not possible.

7 × 108 disintegrations per second. Find the mass of the sample. 03 × l028. –14 42 Encyclopaedia of Atomic Physics Solution: The activity in N atoms is R = λN. 78 × l0–14 per sec. 21 × 1021 atoms. 21 × 1021 atoms in the given sample of U234. 03 × 1023 atoms of U234 = 234 gm. 63 gm. 23. 3 years. 3 years? Solution: The strength, (activity) R of a radioactive source containing N atoms is given by R = λN, where λ is decay constant. 131 per yr. Let N0 be the number of atoms at t = 0 when the strength is 1 milli-curie, and N the number at a time t = 1 year when the strength is R’ (say).

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