JEE Main · 2004 · 24 Apr 2004 · Offline · Chemical Kinetics

The first step is an $\alpha$-emission, so ${}_{92}^{238}\mathrm{M}$ loses a helium nucleus and becomes ${}_{90}^{234}\mathrm{N}$. In the second step two positrons are emitted; each ${\beta }^{+}$ emission turns a proton into a neutron, so the atomic number falls by one while the mass number is unchanged. The daughter $\mathrm{L}$ thus has mass number $234$ and atomic number $90-2=88$, and its neutron count is $234-88=146$. The value $144$ is only the neutron number of the intermediate ${}_{90}^{234}\mathrm{N}$; it quietly forgets the two ${\beta }^{+}$ emissions, which is why it does not describe $\mathrm{L}$.