Question

An electron is confined in a Harmonic potential well with spring constant of 60N/m.

a. What are the energies of its first THREE quantum states?

b. What is the lowest frequency of its emission photon?

c. If the spring constant increases, how will the above values changes? Why?

Answer 1

= sqrt( 60/(9.1*10^-31)) =8.11997943e15 rad/s

= 0.5*6.63*10^-34 /(2pi) *8.11997943e15 =4.28409007e-19 J

or in ev it =4.28409007e-19/ ( 1.6*10^-19) = 2.67755629 eV

E1 = 2.67755629 eV

put n= 2 for enrgy level 2  

E2 = 3/2*6.63*10^-34 /(2pi) *8.11997943e15 =1.28523e-18

or in ev =1.28523e-18/ ( 1.6*10^-19) = 8.0326875 eV

E2 =8.0326875 eV

E3 = 5/2* 6.63*10^-34 /2pi *8.11997943e15 = 2.14204504e-18

or  2.14204504e-18/( 1.6*10^-19) =13.3877815eV   

E3 = 13.3877815eV

b)

jump from 3 to 1

13.3877815eV - 2.67755629 eV =10.71022521 ev

lambda = 6.626*10^-34*3*10^8 /( 10.71022521 *1.6*10^-19 ) =1.15998*10^-7 = 115.998 nm

c.

increase in k will increase  ω and that will increase energy  

and will decrease the wavelength