Question

1.8 gram of hydrogen atom are excited to radiation. the study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atom in second energy level and the rest of in ground state. IP of H is 13.6 EV. calculate (i) number of atom present in 3rd and 2nd energy level (ii) total energy was when all the atoms return to ground state

Answer 1

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

(a) Calculate the number of H atoms

No. of H atoms=1.8g H×1mol H1.008g H×6.022×1023lH atoms1mol H

=1.08×1024lH atoms

(b) Calculate the number of H atoms in 2nd energy level

No. of atoms=0.15×1.08×1024lH atoms=1.61×1023lH atoms

(c) Calculate the number of H atoms in 3rd energy level

No. of atoms=0.27×1.08×1024lH atoms=2.90×1023lH atom

2. Total energy evolved when all atoms return to their ground state

(a) Calculate the value of R

The formula for the energy En of an electron in the nth level of a hydrogen atom is

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣∣aaEn=−Rn2aa∣∣∣−−−−−−−−−−−−−−−−

where

R is the Rydberg energy constant.

The formula for the energy difference between two energy levels nf and ni is

ΔE=R(1n2f−1n2i)

I1 is the energy required to remove an electron from the ground state of the atom.

∴ I1=R(1n21−1∞2)=R(112−0)=R=21.7×10-12lerg

(b) Energy evolved for n=2→1

ΔE2→1=E2−E1=R(112−122)=21.7×10-12lerg×(1−14)=1.63×10-11erg

Total energy=1.61×1023H atoms×1.63×10-11erg1H atom

=2.63×1012lerg

(c) Energy evolved for n=3→1

ΔE3→1=E3−E1=R(112−132)=21.7×10-12lerg×(1−19)=1.93×10-11erg

Total energy=2.90×1023H atoms×1.93×10-11erg1H atom

=5.60×1012lerg

(d) Total energy involved

Total energy=2.63×1012lerg+5.60×1012lerg=8.2×10

12lerg
hope this will help you