JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 13)
Consider an imaginary ion $${ }_{22}^{48} \mathrm{X}^{3-}$$. The nucleus contains '$$a$$'% more neutrons than the number of electrons in the ion. The value of 'a' is _______________. [nearest integer]
Answer
4
Explanation
Number of electrons in $${ }_{22}^{48} X^{3-}$$ is 25 .
Number of neutrons $$=48-22=26$$.
$$\%$$ increase in the number of neutrons over electrons
$$ =\left(\frac{26-25}{25}\right) 100=4 \% $$
$$\therefore a=4$$
Number of neutrons $$=48-22=26$$.
$$\%$$ increase in the number of neutrons over electrons
$$ =\left(\frac{26-25}{25}\right) 100=4 \% $$
$$\therefore a=4$$
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