Question

Suppose a laboratory has a 26-gram sample of polonium-210.The half-life of polonium-210 is about days. a.How many half-lives of polonium-210 occur in 276 days? b.How much polonium is left in the sample after 276 days?

Answer 1

Explanation:

Given:

t1/2 = 138 days

t = 276 days

No = 26 g

t/t1/2 = 276/138

= 2 half-lifes

N(t) = No × (1/2)^(t/t1/2)

= 26 × (1/2)^2

N(276 days) = 6.5 g

Answer 2

a) Two half lives, b)
`m(276) = 6.526\,g`

Explanation:

a) The polonium-210 has a half life of 138.4 days. Therefore, 1.994 half lives have past.

b) Mass decay is described by the following exponential model:

`m(t)=m_(o)\cdot e^{-(t)/(\tau) }`

The time constant for the isotope is:

`\tau = (138.4\,days)/(\ln 2)`

`\tau = 199.669\,days`

The mass of the isotope after 276 days is:

`m(276) = (26\,g)\cdot e^{-(276\,days)/(199.669\,days) }`

`m(276) = 6.526\,g`