Question

Assuming 1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ × 10^–1μg.
[Given : ln 10 = 2.303; log 2 = 0.30]

Answer 1

Correct Answer: 1

To solve the problem of how much of the radioactive element X remains after 100 years, we use the formula for radioactive decay: N(t) = N₀e^–λt, where N(t) is the remaining quantity, N₀ is the initial quantity, λ is the decay constant, and t is time in years. First, compute the decay constant λ using the relationship λ = ln 2 / T₁/₂, where T₁/₂ is the half-life of the element.

Substituting known values, T₁/₂ = 30 years, we get: 

λ = ln 2 / 30.

Given log 2 = 0.30, hence ln 2 = 2.303 × 0.30 = 0.6909.

Therefore, λ = 0.6909 / 30 ≈ 0.02303 per year.

Next, calculate the remaining quantity after 100 years:

N(100) = 1μg × e^{–0.02303 × 100} = 1μg × e^–2.303.

To find e^–2.303, recognize that e^–2.303 = 1 / e^2.303 = 1 / 10 = 0.1.

Therefore, N(100) = 1μg × 0.1 = 0.1μg.

The amount of X remaining is 1.0 × 10^–1 μg, which is within the specified range of the solution, confirming its validity.

Answer 2

\(kt=ln\frac {1}{1−x}\)

\(\frac {0.693}{30}(100)=ln\frac {⁡1}{1−x}\)

\(2.303=2.303\ log⁡\frac {1}{1−x}\)
\(⇒\frac {1}{1−x}=10\)
\(⇒1=10–10X\)
\(⇒X=\frac {9}{10}\)
\(⇒X =0.9\) μg
Amount of X remaining\(=1–X\)
\(=1–0.9\)
\(=0.1\) μg
\(=1×10^{−1}\) μg

So, the answer is \(1\).