The nuclear activity of a radioactive element becomes ${\left( {{1 \over 8}} \right)^{th}}$ of its initial value in 30 years. The half-life of radioactive element is _____________ years.
Answer (integer)
10
Solution
We know, $A = {A_0}{e^{ - \lambda t}}$
<br><br>For half life
<br><br>${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$
<br><br>$\Rightarrow$ ${\lambda {t_{1/2}}}$ = ln 2 .....(1)
<br><br>And when radioactive element becomes ${\left( {{1 \over 8}} \right)^{th}}$ of its initial value in 30 years
<br><br>$${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$$<br><br>$\Rightarrow$ 30$\lambda = 3\ln 2$
<br><br>$\Rightarrow$ $\lambda = {{3\ln 2} \over {30}}$ .....(2)
<br><br>Putting value of $\lambda$ in (1), we get
<br><br>${{3\ln 2} \over {30}} \times {t_{1/2}}$ = ln 2
<br><br>$\Rightarrow$ ${t_{1/2}}$ = 10 years
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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