JEE Main 2026 — Chemistry Physical Chemistry

The energy of an electron in the $n^{th}$nth Bohr orbit of a hydrogen atom is given by $E_n = -R_H \left( \frac{1}{n^2} \right)$En​=−RH​(n21​).

The energy required for excitation from $n=1$n=1 to $n=2$n=2 for a single atom is $\Delta E = E_2 - E_1 = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H$ΔE=E2​−E1​=RH​(121​−221​)=RH​(1−41​)=43​RH​.

Given $R_H = 2.18 \times 10^{-11}$RH​=2.18×10−11 ergs.

Converting ergs to Joules: $1 \text{ erg} = 10^{-7} \text{ J}$1 erg=10−7 J.

So, $R_H = 2.18 \times 10^{-11} \times 10^{-7} = 2.18 \times 10^{-18} \text{ J}$RH​=2.18×10−11×10−7=2.18×10−18 J.

The energy required per atom is $\Delta E = \frac{3}{4} \times 2.18 \times 10^{-18} = 1.635 \times 10^{-18} \text{ J}$ΔE=43​×2.18×10−18=1.635×10−18 J.

To find the energy required per mole of electrons, multiply by Avogadro's number $N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1}$NA​≈6.022×1023 mol−1.

Energy per mole $= 1.635 \times 10^{-18} \times 6.022 \times 10^{23} \text{ J mol}^{-1}$=1.635×10−18×6.022×1023 J mol−1.

Energy per mole $\approx 9.846 \times 10^5 \text{ J mol}^{-1}$≈9.846×105 J mol−1.

Comparing with the given options, $9.835 \times 10^5$9.835×105 is the closest value.