The disintegration of an isotope of sodium,_{ | vedclass

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(D) The given reaction $_{11}Na^{24} 	o _{12}Mg^{24} + _{-1}e^0$ represents $\beta$-decay.In $\beta$-decay,a neutron is converted into a proton,which

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(D) The given reaction $_{11}Na^{24} 	o _{12}Mg^{24} + _{-1}e^0$ represents $\beta$-decay.In $\beta$-decay,a neutron is converted into a proton,which increases the atomic number by $1$ while the mass number remains constant.This process leads to the emission of $\beta$-radiation,helps in achieving a more stable nuclide,and results in a decrease in the neutron $:$ proton ratio.

The disintegration of an isotope of sodium,$_{11}Na^{24} \to _{12}Mg^{24} + _{-1}e^0$,shown is due to

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