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?

1 Answer
Jul 15, 2017

Answer:

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 #"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"#