Which is the correct expression for half-life
${(t)_{1/2}} = \log \,2$
${(t)_{1/2}} = \frac{\lambda }{{\log \,2}}$
${(t)_{1/2}} = \frac{\lambda }{{\log \,{\rm{2}}}}(\,2.303)$
${(t)_{1/2}} = \frac{{2.303\,{\rm{ log\,2}}}}{\lambda }$
In a radioactive sample, ${ }_{10}^a K$ nuclei either decay into stable ${ }_{20}^{* 0} Ca$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable ${ }_{18}^{40}$ Ar muclei with decay constant $0.5 \times 10^{-10}$ per year. Given that in this sample all the stable ${ }_{20}^{\infty 0} Ca$ and ${ }_{15}^{20} Ar$ nuclei are produced by the ${ }_{19}^{* 0} K$ muclei only. In time $t \times 10^{\circ}$ years, if the ratio of the sum of stable ${ }_{30}^{40} Ca$ and ${ }_{15} \operatorname{An}$ nuclei to the radioactive ${ }_{19} K$ muclei is $99$ , the ralue of $t$ will be : [Given $\ln 10=2.3]$
The half-life of radioactive Polonium $(Po)$ is $138.6$ days. For ten lakh Polonium atoms, the number of disintegrations in $24$ hours is
Give a brief explanation about radioactivity.
A freshly prepared radioactive source of half life $2$ hours $30$ minutes emits radiation which is $64$ times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be hours.
The activity of a sample is $ 64 \times 10^{-5}\ Ci$ . Its half-life $3$ days. The activity will become$5 \times {10^{ - 6}}\ Ci$ after ........... $days$