Question

g Consider (12.5 A) micro-grams of a radioactive isotope with a mass number of (78 B) and a half-life of (32.6 C) million years. If energy released in each decay is 32.6 keV, determine the total energy released in joules (J) in 1 (one) year. Give your answer with three significant figures.

Answer 1

Energy released = 18.985 J

Step-by-step explanation:

The exponential decay of radioactive substance is given by -

N(t) = N₀
`e^(-kt)`

where

N₀ = initial quantity

k = decay constant

Half life,
`t_(1/2) = (ln 2)/(k)`

⇒
`k = (ln 2)/(t_(1/2) )`

Given,

N₀ = 12.5 + 3 = 15.5 × 10⁻⁶ gm

`t_(1/2)` = 32.6 + 18 = 50.6 × 10⁶ years

So,

`k = (ln 2)/(50.6 * 10^(6) )` = 1.361 × 10⁻⁸ year⁻¹

Now,

N(1) = 15.5 × 10⁻⁶
`e^{-1.361*10^(-8) *1}`

= 15.49999978904

Now,

Substance decayed = N₀ - N(t)

= 15.5 × 10⁻⁶ - 15.49999978904 × 10⁻⁶

= 21.095 × 10⁻¹⁷ kg

⇒Δm = 21.095 × 10⁻¹⁷ kg

So,

Energy released = Δmc²

= 21.095 × 10⁻¹⁷ × 3 ×10⁸ × 3 × 10⁸

= 189.855 ×10⁻¹

= 18.985 J

⇒Energy released = 18.985 J