Question

Give the mass number and atomic number of elements on the right hand side of the decay process:

²²⁰₈₆Rn → Po + He.

The graph shows how the activity of a sample of radon-220 changes with time. Use the graph to determine its half-life. Calculate the value of decay constant of radon-220.

Answer 1

The given reaction is

²²⁰₈₆Rn → ²¹⁶₈₄Po + ⁴₂He

Mass number of Po = 216, He = 4

Atomic number of Po = 84, He = 2

Half life

From the graph of decay of Rn - 220, we find that at t = 0,

number of nuclei present = N₀ and at t = ∞, N₀ → 0

Since N = N0 e^-λt ...........(i)

As at t = T, N = \(\frac{N_0}{2}\)

∴ \(\frac{N_0}{2}\) = N0e^-λt

or λT = log_e2 = 2.303 log₁₀ 2

or T = \(\frac{2.303 \times 0.3010}{\lambda}\) = \(\frac{0.693}{\lambda}\)

From Eq. (i)

It t = \(\frac{1}{\lambda}\), then

N = N₀ e^{-λ.\(\frac{1}{\lambda} = \frac{N_0}{e}\)}

So decay constant is the time after which the number of radioactive elements reduces to \(\frac{1}{e}\) times the initial number of atoms.