JEE Chemistry 2026 Question with Solution

Given: Statements compare energies of stationary states in one-electron systems.

Find: Which statement gives the correct energy.

For a hydrogen-like species, Bohr's formula is

$$E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \, \text{J/atom}$$

The energy of a bound electron must be negative.

So, options with positive energy are false immediately. Hence Option A and Option B are eliminated.

For Li$^{2+}$ in the third orbit, $Z = 3$ and $n = 3$.

$$E = -2.18 \times 10^{-18} \left( \frac{3^2}{3^2} \right) = -2.18 \times 10^{-18} \, \text{J}$$

This matches the statement in Option C, so it is true.

For H atom in the second orbit, $Z = 1$ and $n = 2$.

$$E = -2.18 \times 10^{-18} \left( \frac{1^2}{2^2} \right) = -0.545 \times 10^{-18} \, \text{J}$$

So Option D is false.

Therefore, the correct option is C.

Given: In hydrogen-like species,

$$E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \, \text{J/atom}$$

Find: The true statement.

If $Z = n$, then

$$\frac{Z^2}{n^2} = 1$$

so the energy becomes exactly

$$-2.18 \times 10^{-18} \, \text{J}$$

which is the energy of the first orbit of hydrogen.

For Li$^{2+}$, $Z = 3$ and for the third orbit $n = 3$, so its energy is exactly $-2.18 \times 10^{-18} \, \text{J}$.

Therefore, the correct option is C.