Question

ergy needed to ionize an atom of Ca when it is in its most stable state is 589.8 kJ mol^-1. However, if an atom of Ca is in a certain low-lying excited state, only 306.9 kJ mol^-1 is needed to ionize it. What is the wavelength of the radiation emitted when an atom of Ca undergoes a transition from this excited state to the ground state? please put answer in nanometers

Answer 1

energy per mol emitted when electron drop from excited state to ground statw

= difference in energy

= 589.8 KJ/mol - 306.9 KJ/mol

= 282.9 KJ/mol

Now find the wavelength of photon.

Given:

Energy of 1 mol = 2.829*10^2 KJ/mol

= 2.829*10^5 J/mol

Find energy of 1 photon first

Energy of 1 photon = energy of 1 mol/Avogadro's number

= 2.829*10^5/(6.022*10^23)

= 4.698*10^-19 J

This is energy of 1 photon

use:

E = h*c/lambda

4.698*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda

lambda = 4.231*10^-7 m

= 423 nm

Answer: 423 nm