1. No. of atoms present in second and third energy level
(a) Calculate the number of $"H"$ atoms
$"No. of H atoms" = 1.8 color(red)(cancel(color(black)("g H"))) × (1 color(red)(cancel(color(black)("mol H"))))/(1.008 color(red)(cancel(color(black)("g H")))) × (6.022 × 10^23color(white)(l) "H atoms")/(1 color(red)(cancel(color(black)("mol H")))$
$= 1.08 × 10^24 color(white)(l)"H atoms"$
(b) Calculate the number of $"H"$ atoms in 2nd energy level
$"No. of atoms" = 0.15 × 1.08 × 10^24 color(white)(l)"H atoms" = 1.61 × 10^23color(white)(l) "H atoms"$
(c) Calculate the number of $"H"$ atoms in 3rd energy level
$"No. of atoms" = 0.27 × 1.08 × 10^24 color(white)(l)"H atoms" = 2.90 × 10^23color(white)(l) "H atoms"$
2. Total energy evolved when all atoms return to their ground state
(a) Calculate the value of $R$
The formula for the energy $E_n$ of an electron in the $n$th level of a hydrogen atom is
$color(blue)(bar(ul(|color(white)(a/a)E_n = -R/n^2color(white)(a/a)|)))" "$
where
$R$ is the Rydberg energy constant.
The formula for the energy difference between two energy levels $n_f$ and $n_i$ is
$ΔE = R(1/n_f ^2- 1/n_i^2)$
$I_1$ is the energy required to remove an electron from the ground state of the atom.
∴ $I_1 = R(1/n_1^2 - 1/∞^2)= R(1/1^2- 0) = R = 21.7 ×10^"-12"color(white)(l) "erg"$
(b) Energy evolved for $n = 2 → 1$
$ΔE_text(2→1) = E_2 - E_1 = R(1/1 ^2- 1/2^2) =21.7 × 10^"-12"color(white)(l) "erg" × (1 - 1/4) = 1.63 × 10^"-11" "erg"$
$"Total energy" = 1.61 × 10^23 color(red)(cancel(color(black)("H atoms"))) × (1.63 × 10^"-11" "erg")/(1 color(red)(cancel(color(black)("H atom"))))$
$= 2.63 × 10^12color(white)(l) "erg"$
(c) Energy evolved for $n = 3 → 1$
$ΔE_text(3→1) = E_3 - E_1 = R(1/1 ^2- 1/3^2) =21.7 × 10^"-12"color(white)(l) "erg" × (1 - 1/9) = 1.93 × 10^"-11" "erg"$
$"Total energy" = 2.90 × 10^23 color(red)(cancel(color(black)("H atoms"))) × (1.93 × 10^"-11" "erg")/(1 color(red)(cancel(color(black)("H atom"))))$
$= 5.60 × 10^12color(white)(l) "erg"$
(d) Total energy involved
$"Total energy" = 2.63 × 10^12color(white)(l) "erg" + 5.60 × 10^12color(white)(l) "erg" = 8.2 × 10^12color(white)(l) "erg"$
