# 1.8 grams of atoms of hydrogen are excited to radiations. the study of spectra indicates that 27% of atoms are in the third energy level and 15% of atoms ate in the second energy level and rest in the ground state. If the I.P. of H is 21.7*10^(-12)erg?

Jul 15, 2017

Warning! Long Answer. 1. 1.61 × 10^23 and 2.90 × 10^23; 2. 8.2 × 10^12color(white)(l) "erg"

#### Explanation:

1. No. of atoms present in second and third energy level

(a) Calculate the number of $\text{H}$ 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 $\text{H}$ atoms in 2nd energy level

$\text{No. of atoms" = 0.15 × 1.08 × 10^24 color(white)(l)"H atoms" = 1.61 × 10^23color(white)(l) "H atoms}$

(c) Calculate the number of $\text{H}$ atoms in 3rd energy level

$\text{No. of atoms" = 0.27 × 1.08 × 10^24 color(white)(l)"H atoms" = 2.90 × 10^23color(white)(l) "H atoms}$

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {E}_{n} = - \frac{R}{n} ^ 2 \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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

$\text{Total energy" = 2.63 × 10^12color(white)(l) "erg" + 5.60 × 10^12color(white)(l) "erg" = 8.2 × 10^12color(white)(l) "erg}$