Question

The energy needed to ionize an atom of Zn when it is in its most stable state is 906.4 kJ mol. However, if an atom of Zn is in a certain low-lying excited state, only 347.2 kJ mol is needed to ionize it. What is the wavelength of the radiation emitted when an atom of Zn undergoes a transition from this excited state to the ground state? nm Submit Answer 5 question attempts remaining

Answer 1

Energy emitted when transition happens from excited state to ground state,
E = 906.4 KJ/mol - 347.2 KJ/mol
= 559.2 KJ/mol

Given:
Energy of 1 mol = 5.592*10^2 KJ/mol
= 5.592*10^5 J/mol

Find energy of 1 photon first
Energy of 1 photon = energy of 1 mol/Avogadro's number
= 5.592*10^5/(6.022*10^23)
= 9.286*10^-19 J
This is energy of 1 photon

use:
E = h*c/lambda
9.286*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 2.141*10^-7 m
lambda = 214.1 nm
Answer: 214.1 nm