JEE Chemistry 2023 Question with Solution

Given: The radius of the first Bohr orbit of hydrogen atom is $0.6 \, \text{Å}$.

Find: The radius of the third Bohr orbit of $\mathrm{He}^{+}$ in picometer.

The radius of the $n$-th Bohr orbit is given by:

$$r_n = r_1 \frac{n^2}{Z}$$

For $n = 3$ and $Z = 2$:

$$r_{\mathrm{He}^{+}} = 0.6 \times \frac{3^2}{2} = 2.7 \, \text{Å}$$

Converting to picometers:

$$2.7 \, \text{Å} = 270 \, \text{pm}$$

Therefore, the required numerical value is $270$.

Given: For hydrogen, $r_1 = 0.6 \, \text{Å}$.

Find: Radius of the third orbit of $\mathrm{He}^{+}$.

For a hydrogen-like species, orbit radius varies as:

$$r_n \propto \frac{n^2}{Z}$$

So for $\mathrm{He}^{+}$, use $n = 3$ and nuclear charge $Z = 2$:

$$r_3 = 0.6 \times \frac{9}{2}$$ $$r_3 = 0.6 \times 4.5 = 2.7 \, \text{Å}$$

Now convert angstrom to picometer:

$$1 \, \text{Å} = 100 \, \text{pm}$$

Hence,

$$2.7 \, \text{Å} = 270 \, \text{pm}$$

So the answer is $270$.