JEE MAIN - Chemistry (2025 - 22nd January Morning Shift - No. 19)
Radius of the first excited state of Helium ion is given as : $\mathrm{a}_0 \rightarrow$ radius of first stationary state of hydrogen atom.
$\mathrm{r}=\frac{\mathrm{a}_0}{4}$
$\mathrm{r}=2 \mathrm{a}_0$
$\mathrm{r}=4 \mathrm{a}_0$
$\mathrm{r=\frac{a_0}{2}}$
Explanation
$$ r_n = \frac{n^2\, a_0}{Z} $$
For the helium ion, which is hydrogen-like with a nuclear charge of $$ Z = 2 $$, the first excited state corresponds to $$ n = 2 $$. Substituting these values:
$$ r_2 = \frac{(2)^2\, a_0}{2} = \frac{4\, a_0}{2} = 2\, a_0 $$
Thus, the radius of the first excited state of the helium ion is $$ 2\, a_0 $$.
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