1) The wavelength of a Ka line is 62 pm. Find the energy difference between the
ID: 1451358 • Letter: 1
Question
1) The wavelength of a Ka line is 62 pm. Find the energy difference between the ground state and the first excited level.
2) Consider an atom with a ground state of -8.50 eV and two excited states, a and b, at Ea = -5.70 eV and Eb = - 3.50 eV. The atom is excited from the ground state to b by ultraviolet light.
a) Find the frequency of the ultraviolet light.
b) If the atom de-excites in two steps (b a and a ground state), find the wavelengths of the light emitted.
please show all work and formulas used.
Explanation / Answer
1) the energy difference between the ground state and the first excited level = Energy of emitted photon
= h*c/lamda
= 6.626*10^-34*3*10^8/(62*10^-12)
= 3.2*10^-15 J
= 3.2*10^-15/(1.6*10^-19) eV
= 2*10^4 eV
2) Energy of emited phton, E = Eb - Ea
h*f = Eb - Ea
h*f = -3.5 - (-8.5)
h*f = 5 eV
f = 5*1.6*10^-19/(6.626*10^-34)
= 1.21*10^15 hz
b) for transition b --> a
E = Eb - Ea
h*c/lamda = EB - Ea
==> lamda = h*c/(Eb - Ea)
= 6.626*10^-34*3*10^8/( (-3.5 + 5.7)*1.6*10^-19)
= 5.65*10^-7 m or 565 nm
for transition a --> ground
E = Ea - Eg
h*c/lamda = Ea - Eg
==> lamda = h*c/(Ea - Eg)
= 6.626*10^-34*3*10^8/( (-5.7+8.5)*1.6*10^-19)
= 4.44*10^-7 m or 444 nm
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