1.

Various rules of thumb have seen proposed by the scientific community to expalin the mode of radioactive decay by various radioisotopes. One of the major rules is called the `n//p` ratio. If all the known isotopes of the elemnts are plotted on a graph of number of neutrons (n) versus number of protons (p), it is observed that all isotopes lying outside of a 'stable' `n//p` ratio region are radioactive as shown f The graph exhibits straight line behaviour with unit slope up to `p=25`. Above `p=25`, tgose isotopes with `n//p` ratios lying above the stable region usually undergo beta decay. Very heavy isotopes `(pgt83)` are unstable because of their relativley large nuclei and they undergo alpha decay. Gamma ray emission does not involve the release of a particle. It represnts a change in an atom from a higher energy level to a lower energy level. `Th-230` undergoes a series of radioactive decay processes resulting in `Bi-214` being the final product. What was the sequence of the processes that occured?A. `alpha,alpha,alpha,g,beta`B. `alpha,alpha,alpha,alpha,beta`C. `alpha,alpha,beta,beta`D. `alpha,beta,beta,beta,gamma`

Answer» Correct Answer - b
`._(90)^(230)Th rarr ._(83)^(214)Bi + x._(1)^(0)e^(bar) +y._(2)^(4)He^(2+)+z ("gamma ray")`
Since the sum of the atomic numbers and mass numbers on either side of the equation must be equal (matter cannot be created or destroyed),we get
`90=83-x+2y` and `230=214+4y`
Solving, we get
`y=4` and `x=1`
The order of the reactions is irrelevant (i.e., `alpha, beta`,...) Since gamma rays have no atomic number or mass number, the value of z does not affect this particular calculation.


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