The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then :

$\left(t_{1 / 2}\right)_{x}=(\tau)_{y}$

$\Rightarrow \frac{\ell \mathrm{n} 2}{\lambda_{\mathrm{x}}}=\frac{1}{\lambda_{\mathrm{y}}} \Rightarrow \lambda_{\mathrm{x}}=0.693 \lambda_{\mathrm{y}}$

Also initially $\mathrm{N}_{\mathrm{x}}=\mathrm{N}_{\mathrm{y}}=\mathrm{N}_{0}$

Activity $\mathrm{A}=\lambda \mathrm{N}$

As $\lambda_{x}<\lambda_{y} \Rightarrow A_{x}

$\Rightarrow y$ will decay faster than $\mathrm{x}$

Option (2)