Question

A radio-active elements has half-life of 15 years. What is the fraction that will decay in 30 years?

• A. 0.25
• B. 0.5
• C. 0.75
• D. 0.85

Answer 1

The Correct Option is C

The fraction of a radioactive element that decays over a given period of time is given by the formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \] Where: 
\(N(t)\) is the amount of the substance remaining after time \(t\),
\(N_0\) is the initial amount of the substance,
\(T_{1/2}\) is the half-life of the substance.
The fraction that decays is given by: \[ \text{Fraction decayed} = 1 - \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
Given: 
\(T_{1/2} = 15 \, \text{years}\),
\(t = 30 \, \text{years}\). Substitute the values: \[ \text{Fraction decayed} = 1 - \left(\frac{1}{2}\right)^{\frac{30}{15}} = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = 0.75 \] Thus, the fraction that will decay in 30 years is 0.75.

The correct answer is (C) : 0.75.

Answer 2

The decay of a radioactive element follows an exponential decay law, and the fraction remaining after a time \( t \) is given by the formula: \[ \frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \] Where:
\( N(t) \) is the amount remaining after time \( t \),
\( N_0 \) is the initial amount,
\( T_{1/2} \) is the half-life,
\( t \) is the time elapsed.

We are given:
\( T_{1/2} = 15 \, \text{years} \),
\( t = 30 \, \text{years} \).

Now, the fraction remaining after 30 years is: \[ \frac{N(30)}{N_0} = \left(\frac{1}{2}\right)^{\frac{30}{15}} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25 \] So, the fraction that decays is: \[ \text{Fraction decayed} = 1 - 0.25 = 0.75 \] Therefore, the fraction that will decay in 30 years is 0.75. Hence, the correct answer is (C).