Question

The half-life period of ₈₄P²¹⁰ is 140 days. In how many days 1g of this isotope is reduced 1 to 0.25 g-
1. 120 days
2. 140 days
3. 380 days
4. 280 days

Answer 1

Correct Answer - Option 4 : 280 days

Concept:

• Radioactivity: The atoms having a number of neutrons much more than proton are unstable. The nuclei of such atoms exhibit radioactivity.
  • An example of Such an Atom is U - 238, where the number of Neutrons is 146, and the number of protons is 92.
  • Radium is an important radioactive atom.
  • Radioactive decay: The spontaneous breakdown of such an unstable atomic nucleus causes radioactivity.
  • The process of radioactive decay as a function of time is represented by

\(ln (\frac{N}{N_0}) = - λ t\)

N is the number of radioactive nuclei present in a sample, No is the number of radioactive nuclei present in the sample at t = 0, λ is disintegration constant.

• Disintegration constant: The disintegration constant is the unique constant related to the radioactive nature of the given nucleus. 
• Half-Life Period: The time in which the number of nuclei reached to half of its initial value is called the half-life period.

It is the time period when N = N0 / 2

The half - Life period is given as

\(ln (\frac{N_0 /2}{N_0}) = - λ t_{1/2}\)

⇒ \(t_{1/2} = \frac{ln2}{λ}\)

\(\implies \lambda= \frac{ln2}{t_{1/2}}\)

Calculations:

Given half-life t _1/2 = 140 days

Disintegration constant 

\( \lambda= \frac{ln2}{140}\)

Initial Amount present N₀ = 1 gm

Amount present now  N = 0. 25 gm

\(\frac{N}{N_0} = \frac{1}{4}\)

So, we have 

\(ln (\frac{N}{N_0}) = - λ t\)

\(\implies ln ( \frac{1}{4} ) = - \frac{ln2}{140} × t\)

 \(\implies - ln (4 ) = - \frac{ln2}{140} × t\)

\(\implies t = \frac{ln4}{ln2} × 140\)

⇒ t = 2 × 140 = 280 days

So, 280 days is the correct answer.

Properties of Logarithms used

\(log (\frac{1}{a}) = - log a\)

\(log (a^n) = 2 log a\)