QJEEM23Ch63ad: JEE Main 2023 Chemistry - Class 11 JEEM23Ch63ad Solution: The electron in the orbit of is excited to orbit using the…

The energy levels of a hydrogen-like atom (such as Li²⁺ in this case) can be calculated using the formula:

$E = -R_H \cdot \frac{Z^2}{n^2}$

where:

• $E$ is the energy of the electron,
• $R_H$ is the Rydberg constant (in joules),
• $Z$ is the atomic number, and
• $n$ is the principal quantum number.

The change in energy (∆E) between two energy levels is given by the difference in the energies of the two levels. In this case, the electron is excited from the nth level to the (n+1)th level, and the energy of the radiation is given as $1.47 \times 10^{-17}$ J. Therefore:

$\Delta E = E_{n+1} - E_n = 1.47 \times 10^{-17} \mathrm{~J}$

Substituting the formula for $E_{n}$ and $E_{n+1}$ :

$\Delta E = -R_H \cdot \frac{Z^2}{(n+1)^2} - \left(-R_H \cdot \frac{Z^2}{n^2}\right)$

Simplifying this, we get:

$\Delta E = R_H \cdot Z^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

We know that $Z=3$ for $\mathrm{Li}^{2+}$ , $R_H = 2.18 \times 10^{-18} \mathrm{~J}$ and $\Delta E = 1.47 \times 10^{-17} \mathrm{~J}$ .

We substitute the given values into the formula

$1.47 \times 10^{-17} \mathrm{~J} = 2.18 \times 10^{-18} \mathrm{~J} \times 3^2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

Simplify to

$\frac{1.47}{1.96} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$

Solving this quadratic equation for n, we get two possible solutions. However, we disregard the negative solution since n (the principal quantum number) can't be negative. So, the only valid solution is n=1.

Thus, the value of n is 1.