Question

A radioactive substance of half-life 69.3 days is kept in a container. The time in which 80% of the substance will disintegrate will be ______

[take ln(5) = 1.61]

Options

• 1.61 days

• 16.1 days

• 161 days

• 1610 days

Answer 1

A radioactive substance of half-life 69.3 days is kept in a container. The time in which 80% of the substance will disintegrate will be 161 days.

Explanation:

T = 69.3 days

∴ λ = `0.693/"T" = 0.693/69.3 = 10^-2` per day

N = 20% of N₀ = 0.2 N₀

Now, N = `N_0e^{-lambdat}`, which gives

`N_0/N = e^{lambdat}` or ln`(N_0/N) = lambdat`

∴ t = `ln(N_0/N)/lambda`

= `ln(5)/10^-2 = 1.61/10^-2 = 161` days