A radioactive nucleus decays by two different processes. The half life for the first process is 10 s and that for the second is 100 s. The effective half life of the nucleus is close to :
Solution
T<sub>1</sub> = 10 sec
<br><br>$\lambda$<sub>1</sub> = ${{\ln 2} \over {{T_1}}}$
<br><br>T<sub>2</sub> = 100 sec
<br><br>$\lambda$<sub>2</sub> = ${{\ln 2} \over {{T_2}}}$
<br><br>$\lambda$<sub>eq</sub> = ${{\ln 2} \over {{T_{eq}}}}$
<br><br>We know,
<br><br>$\lambda$<sub>eq</sub> = $\lambda$<sub>1</sub> + $\lambda$<sub>2</sub>
<br><br>$\Rightarrow$ ${{\ln 2} \over {{T_{eq}}}}$ = ${{\ln 2} \over {{T_1}}}$ + ${{\ln 2} \over {{T_2}}}$
<br><br>$\Rightarrow$ ${1 \over {{T_{eq}}}}$ = ${1 \over {10}} + {1 \over {100}}$
<br><br>$\Rightarrow$ T<sub>eq</sub> = ${{100} \over {11}}$ = 9 sec
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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