The half-life of \(^{32}\)P is 14.26 d. What percentage of \(^{32}\)P

1.	The half-life of \(^{32}\)P is 14.26 d. What percentage of \(^{32}\)P sample will remain after 40 d?
Answer» Given: t_1/2 = 14.26 d, N₀ = 100, t = 40 d To find: Percentage of \(^{32}\)P sample remaining after 40 d Formula: i. t_1/2 = \(\frac{0.693}{\lambda}\) ii. \(\lambda\) = \(\frac{2.303}{t}\) log₁₀ \((\frac{N_0}{N})\) Calculation: \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{14.26\,d}\) = 0.0486 d^-1 Now, \(\lambda\) = \(\frac{2.303}{t}\) log₁₀ \((\frac{N_0}{N})\) Hence, log₁₀ \(\frac{N_0}{N}\) = \(\frac{\lambda t}{2.303}\) = \(\frac{0.0486\,d^{-1}\times 40\,d}{2.303}\) = 0.8441 Taking antilog of both sides we get, \(\frac{N_0}{N}\) = antilog (0.8441) = 6.984 ∴ \(\frac{100}{N}\) = 6.984 or N = \(\frac{100}{6.984}\) = 14.32% % \(^{32}P\) that remains after 40 d is 14.32%.

The half-life of \(^{32}\)P is 14.26 d. What percentage of \(^{32}\)P sample will remain after 40 d?

Answer»

Given: t_1/2 = 14.26 d,

N₀ = 100,

t = 40 d

To find: Percentage of \(^{32}\)P sample remaining after 40 d

Formula:

i. t_1/2 = \(\frac{0.693}{\lambda}\)

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

Calculation: \(\lambda\) = \(\frac{0.693}{t_{1/2}}\) = \(\frac{0.693}{14.26\,d}\) = 0.0486 d^-1

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

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

= \(\frac{0.0486\,d^{-1}\times 40\,d}{2.303}\) = 0.8441

Taking antilog of both sides we get,

\(\frac{N_0}{N}\) = antilog (0.8441) = 6.984

∴ \(\frac{100}{N}\) = 6.984 or N = \(\frac{100}{6.984}\) = 14.32%

% \(^{32}P\) that remains after 40 d is 14.32%.