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[KuMoRuSaKi_ICHI]

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1) 21083Bi ---------------> 21084Po + 0-1e

2) 5526Fe + 0-1e -------------> 5525Mn

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Ans 1 :

The missing nuclide atomic number can be determined by adding the atomic numbers of the both the product species and likewise the mass numbers can be determined by adding the mass numbers .

So here :

atomic number = 2 + 81 = 83

mass number = 4 + 206 = 210

So the nuclide symbol will be : 83210Bi

Ans 2 :

When a nuclide undergoes a positron emission , a positron is released alongwith the product nuclide that has the same mass number as the reactant nuclide but one atomic number lesser.

So the name of the product nuclide here will be : Carbon -13

The symbol of the product nuclide is : 613C

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N = Noe-λt , where N is quantity of sample still remained(not decayed), No is initial quantity of the sample, λ is the decay constant and t is time taken.

⇒ N/No = e-λt

Taking log with base e [ln] both the sides :-

⇒ ln{N/No} = ln{e-λt }

⇒ ln{N/No} = -λt

⇒ ln{No /N} = λt

For sample of Uranium-238 to fall to 25.2 percent of its original value,t = 8.97*109 years, N = 0.252No

⇒ ln {No / (0.252No) = λ*8.97*109

⇒ ln{1/0.252} = λ*(8.97*109)

⇒ λ = ln{3.9682}/8.97*109 =1.5366*10-10 (year)-1

Now for half life, t=t1/2, N=No/2 and λ =1.5366*10-10 (year)-1

⇒ ln {No/(No/2)} = (1.5366*10-10)*t1/2

⇒ t1/2 = ln{2}/(1.5366*10-10) = 4.51*109 years

Therefore half life of Uranium-238 is 4.51*109 years.

 

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