Question

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?

Answer 1

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"`