Question

The half-life of Calcium-47 is 3.91 x 105 seconds. A sample contains 4.11 x 1076 nuclei. How much of the sample is left after 7.82 x 10 seconds?
2.57 x 1015 nuclei
5.14 x 1015 nuclei
1.03 x 1016 nuclei
2.06 x 1016 nuclei

Answer 1

Remember the formula

$\\ \rm\Rrightarrow A(t)=A_o(0.5)^{\frac{t}{h}}$

• h is half life

$\\ \rm\Rrightarrow A(7.82\times 10)=A(78.2)$

$\\ \rm\Rrightarrow 4.11\times 10^{76}(0.5)^{\frac{78.2}{3.91\times 10^{5}}}$

$\\ \rm\Rrightarrow 4.11\times 10^{76}(0.5)^{20\times 10^{-5}}$

$\\ \rm\Rrightarrow 4.11\times 10^{76}(0.5)^{0.0002}$

$\\ \rm\Rrightarrow 4.11\times 10^{76}(0.998)$

$\\ \rm\Rrightarrow 4.10178\times 10^{76}$

Why no options matched?

That time given in end i.e 7.82×10

• 10 should contain some power to match our options .

Answer 2

The number of nuclei  of calcium-47 left after 7.82 x 10⁵s is 1.03 x 10¹⁶.

This can be found out using half life and laws of radioactive disintegration.

What are the laws of radioactive disintegration?

The Law of radioactive disintegration states  the number of  radioactive decay per unit time is proportional to the number of nuclei present in the sample initially.

So,  ΔN/Δt = λN

Where, ΔN = number of radioactive nuclei decayed

Δt = time taken for decay

N = No of nuclei present initially

λ= decay constant

What is Half life?

Half life is the time taken for the radioactive nuclei to reach half of its initial value.

Half life is given by

 $t_{1/2}$ =  λ/t

Given here,  $t_{1/2}$ =  3/91 x 10.⁵s

N = 4.11 x 10¹⁶.

t = 7.82 x 10.⁵s

We see  here that  t = 2($t_{1/2}$)

Now, number of radioactive nuclei left undecayed in  one half life is N/2

So number of nuclei left undecayed in 2 half lives will be 1/2(N/2) = N/4

So number of nuclei left in the sample will be N/4 = (4.11 x 10¹⁶.)/4

So Number of nuclei left in the sample = 1.03 x 10¹⁶.

Hence 1.03 x 10¹⁶is the correct answer.

To know more on Half life here

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