Question

Although not visible through your spectroscope, photons of both higher and lower energy are being emitted from the hydrogen discharge tube. Calculate one transition that would lead to the emission of a UV (ultraviolet) photon. (UV light has a wavelengths between approximately 4 and 400 nm). Of an IR (infared) photon? (IR light has wavelengths between approximately 700 and 700,000 nm).

Answer 1

You have to use the Rydberg formula:

1/λ = R ( 1/n_f² – 1/n_i² )

Where the Rydberg constant R is 1.097x10⁷ m^-1 = 0.010972 nm^-1

You may write

λ= 1/ [( 1/n_f² – 1/n_i² )R ]

and also

( 1/n_f² – 1/n_i² ) =1/λR

a). For an UV line λ < 400 nm and

1/λR > 1/ (400 x 0.010972 nm^-1) or 1/λR > 0.227

and also                                                          

1/n_f² – 1/n_i² > 0.227

If n_f = 1 , 1/n_i² < 0.773 or n_i² > 1/0.773 or n_i > 1.3 ( read as n_i >2)

Any transition to n_f = 1 is in UV.

   λ= 1/ [( 1/n_f² – 1/n_i² )R ]

E.g., if n_i = 2 , λ= 1/ [(1/1 -1/4)x 0.010972 nm^-1) ] = 121.5 nm

b). For an IR transition λ > 700nm and

1/λR < 1/ (700 x 0.010972 nm^-1) or 1/λR < 0.13

and also                                                          

1/n_f² – 1/n_i² < 0.130

If n_f = 3 , and n_i >3 the condition is satisfied

Any transition to n_f = 3 is in IR.

   λ= 1/ [( 1/n_f² – 1/n_i² )R ]

E.g., if n_i = 4 , λ= 1/ [(1/9 -1/16)x 0.010972 nm^-1) ] = 1879.2 nm