- ADoes not change
- ✓Increases by $1$
- CDecreases by $1$
- DIncreases by $2$
When an atom undergoes $\beta-$decay the atomic number increases by $1.$
When an atom undergoes $\beta-$decay, one of the neutrons breaks into one proton and one electron. The resultant electron is then ejected out of the nucleus and this is called as the $\beta$ particle.
While the resultant proton stays inside the nucleus which results in increase of atomic number by $1$, whereas the atomic mass remains invariant.
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Frequency in $Hz$ Phase lead of $N$ over $M$ in radians

In the given nuclear reaction, the approximate amount of energy released will be $.....\,MeV$
$\text { [Given, mass of }{ }_{92}^{238} A =238.05079 \times 931.5\,MeV / c ^2$
$\text { mass of }{ }_{90}^{234} B =234.04363 \times 931.5\,MeV / c ^2$
$\text { mass of } \left.{ }_2^4 D =4.00260 \times 931.5\,MeV / c ^2\right]$