QJEEAdv23Ch64c4: JEE Advanced 2023 Chemistry - Class 11 JEEAdv23Ch64c4 Solution: For , a transition takes place from the orbit of radius to the…

1. The radius of the nth orbit in a single-electron system is given by the equation:

$r = \frac{52.9 \times n^2}{Z}$

where $r$ is in pm, $n$ is the principal quantum number, and $Z$ is the atomic number.

2. For a $\mathrm{He}^{+}$ ion, $Z=2$ .

3. We are given the initial radius $r_2 = 105.8$ pm and the final radius $r_1 = 26.45$ pm.

4. We can substitute these radii into the equation for $r$ to find the corresponding quantum numbers.

For $r_2 = 105.8$ pm, we get :

$105.8 = \frac{52.9 \times n_2^2}{2}$

which gives $n_2 = 2$ .

Similarly, for $r_1 = 26.45$ pm, we get :

$26.45 = \frac{52.9 \times n_1^2}{2}$

which gives $n_1 = 1$ .

5. Therefore, the transition is from $n_2 = 2$ to $n_1 = 1$ .

6. The energy difference during this transition is equal to the energy of the emitted photon, which is given by :

$E = \frac{hc}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where $h$ is the Planck's constant, $c$ is the speed of light, $R_H$ is the Rydberg constant, and $\lambda$ is the wavelength of the photon.

7. Substituting all known values into this equation gives :

$\frac{6.6 \times 10^{-34} \, \mathrm{J} \, \mathrm{s} \times 3 \times 10^8 \, \mathrm{m/s}}{\lambda} = 2.2 \times 10^{-18} \, \mathrm{J} \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$

Solving this equation yields $\lambda = 30 \times 10^{-9}$ m, or 30 nm.

8. So, the wavelength of the emitted photon during the transition is 30 nm.