The half-life of \(^{209}\)Po is 102 y. How much of 1 mg sample of pol

1.	The half-life of \(^{209}\)Po is 102 y. How much of 1 mg sample of polonium decays in 62 y?
Answer» Given: t_1/2 = 102y, t = 62 y, N₀ = 1 mg To find: Amount of polonium that decayed in 62 y Formula: i. t_1/2 = \(\frac{0.693}{\lambda}\) ii.\(\lambda\) = \(\frac{2.303}t\) log₁₀\((\frac{N_0}N)\) Calculation: \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{102y}\) = 6.794 x 10^-3 y^-1 \(\lambda\) = \(\frac{2.303}{t}\) log₁₀\((\frac{N_0}N)\) Hence, log₁₀\((\frac{N_0}N)\) = \(\frac{\lambda t}{2.303}\) = \(\frac{6.794\times 10^{-3}y^{-1}\times 62\,y}{2.303}\) = 0.1829 Taking antilog of both sides we get, \(\frac{N_0}N\) = antilog (0.1829) = 1.524 N = \(\frac{N_0}{1.524}\) = \(\frac{1\,mg}{1.524}\) = 0.656 mg N is the amount that remains after 62 y. Hence, the amount decayed in 62 y = 1 mg – 0.656 mg = 0.344 mg The amount decayed in 62 y is 0.344 mg

The half-life of \(^{209}\)Po is 102 y. How much of 1 mg sample of polonium decays in 62 y?

Answer»

Given: t_1/2 = 102y,

t = 62 y,

N₀ = 1 mg

To find: Amount of polonium that decayed in 62 y

Formula:

i. t_1/2 = \(\frac{0.693}{\lambda}\)
ii.\(\lambda\) = \(\frac{2.303}t\) log₁₀\((\frac{N_0}N)\)

Calculation: \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{102y}\) = 6.794 x 10^-3 y^-1

\(\lambda\) = \(\frac{2.303}{t}\) log₁₀\((\frac{N_0}N)\)

Hence, log₁₀\((\frac{N_0}N)\) = \(\frac{\lambda t}{2.303}\)

= \(\frac{6.794\times 10^{-3}y^{-1}\times 62\,y}{2.303}\) = 0.1829

Taking antilog of both sides we get, \(\frac{N_0}N\) = antilog (0.1829) = 1.524

N = \(\frac{N_0}{1.524}\) = \(\frac{1\,mg}{1.524}\) = 0.656 mg

N is the amount that remains after 62 y. Hence, the amount decayed in 62 y = 1 mg – 0.656 mg = 0.344 mg

The amount decayed in 62 y is 0.344 mg