Question

A large number of ground state electrons in a gas of hydrogen atoms are excited to the fourth excited state.

a. How many energy did each electron gain?

b. How many visible photos would be emitted by these electrons as they return to the ground state?

c. What is the wavelength (in nm) of the least energetic visible photon?

Answer 1

Energy absorbed by electron from going to n1 to n2 is E=13.6[1/(n1)^2 -1/(n2)^2]

So for n1=1 and n2=5

E =13.6(1-1/25) =13.056 eV

Possible transition from n=5 are to n'=4,3,2,1

For 5 to 4 Energy emitted =13.6[1/16-1/25] =0.306eV =>wave length=hc/E= 4.06 micro meter

For 5 to 3 Energy emitted =13.6[1/9-1/25] =0.97eV =>wave length=hc/E= 1.28 micro meter

For 5 to 2 Energy emitted =13.6[1/4-1/25] =2.85eV =>wave length=hc/E= 436 nm

For 5 to 1 Energy emitted =13.6[1-1/25] =13.056eV =>wave length=hc/E= 95 nm nm

For 4 to 3 Energy emitted =13.6[1/9-1/16] =0.66eV =>wave length=hc/E= 1.88 micro meter

For 4 to 2 Energy emitted =13.6[1/4-1/16] =2.55eV =>wave length=hc/E= 487 nm

For 4 to 1 Energy emitted =13.6[1-1/16] =12.75eV =>wave length=hc/E= 97.4 nm

For 3 to 2 Energy emitted =13.6[1/4-1/9] =1.89eV =>wave length=hc/E= 657 nm

For 3 to 1 Energy emitted =13.6[1-1/9] =12.09eV =>wave length=hc/E=102 nm

For 2 to 1 Energy emitted =13.6[1-1/4] =10.2eV =>wave length=hc/E=121.8 nmnm

So 3 wavlelengths in visible range

wave length of least energitic photon is 4.06 micro meter