Question

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was originally present?

Answer 1

it's 11.2, other guy prob made a typo is all

Answer 2

Answer: The initial amount of Uranium-232 present is 11.3 grams.

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=68.9yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{68.9}=0.0101yr^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.0101yr^{-1}$

t = time taken for decay process = 206.7 yrs

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

$0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}$

$[A_o]=11.3g$

Hence, the initial amount of Uranium-232 present is 11.3 grams.