Although not visible through your spectroscope, photons of both higher and lower

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). 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). 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).

Explanation / Answer

You have to use the Rydberg formula:

1/ = R ( 1/nf2 – 1/ni2 )

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

You may write

= 1/ [( 1/nf2 – 1/ni2 )R ]

and also

( 1/nf2 – 1/ni2 ) =1/R

a). For an UV line < 400 nm and

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

and also                                                          

1/nf2 – 1/ni2 > 0.227

If nf = 1 , 1/ni2 < 0.773 or ni2 > 1/0.773 or ni > 1.3 ( read as ni >2)

Any transition to nf = 1 is in UV.

   = 1/ [( 1/nf2 – 1/ni2 )R ]

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

b). For an IR transition > 700nm and

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

and also                                                          

1/nf2 – 1/ni2 < 0.130

If nf = 3 , and ni >3 the condition is satisfied

Any transition to nf = 3 is in IR.

   = 1/ [( 1/nf2 – 1/ni2 )R ]

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.