Question

A radioactive material decays by simultaneous emissions of two particles with half-lives of 1400 years and 700 years respectively. What will be the time after which one-third of the material remains? (Take In3 = 1.1)

Options

• 1110 years

• 340 years

• 740 years

• 700 years

Answer 1

740 years

Explanation:

A into B has a half-life of 1400 years.

A and C have half-lives of 700 years.

\[\ce{A ->[λ_1]B}\]     \[\ce{A->[λ_2]C}\]

`[T_(1"/"2)]_(AB) = (log_e2)/(lambda_1)`

`[T_(1"/"2)]_(AC) = (log_e2)/(lambda_2)`

`lambda_"net" = lambda_1 + lambda_2 = (In2)/1400 + (In2)/700`

= `In2[3/1400]`

Use this phrase,

`N = N_0e^{-lambdat}`

`N_0/3 = N_0e^{-lambda_"net"t}`

`log_e3 = lambda_"net"t`

⇒ `1.1 = 0.693[3/1400]t`

t = 740 years (approx)