Question

The hydrogen atom has an emission at 1875 nm. What transition is responsible for this spectral line? What region of the electromagnetic spectrum would it appear? (7 pts)

Answer 1

From the Rydberg equation

1/λ= R (1/n12-1/n22)

λ is he wavelength of the emitted photon

R is Rydberg constant = 1.097 x 107 /m

n1 = Principal quantum number of lower energy level

n2 = principal quantum number of higher energy level

If n1 < n2 then we don’t get negative value

1875 nm x 1m / 10-9 nm = 1875 x 10-9 m

(1/n12-1/n22) = 1/R x 1/λ

Substitute R and λ in the above equation

(1/n12-1/n22) = (1/1.097 x 107 m-1) x (1/1875 x 10-9 m)

                       = 0.912 x 10-7 x 5.333 x 10-4 x 109

                       = 48.62 x 10-2 =0.486 =0.49

(1/n12-1/n22) = 0.49

n22 - n12/ n12 n22 = 0.49 = 49/100

n22 - n12 = 49

n12 n22 = 100

From the above two equations

n1 n2 = sqrt(100) = 10

n1 = 10/n2

n22 - (10/n2)2 =49

n22 - 100/n22 = 49

n24 -49 n22 -100 = 0

let n22 = X

then X2-49X -100 = 0

X = -(-49) ± [( -49)2-4(1)(-100)]1/2 /2 x 1

   = (49 ± (22401)1/2 )/2

= (49 ± 149.67) / 2

X = 99. 34 , -50.335

n22 = X = 99.34

n2 = (99.34)1/2 = 9.967 = ~10

n1 = 10/n2 = 10/10 = 1

The transition takes place from n = 10 to n= 1

It comes in Infrared region

Infrared region ranges from 25µm to 2.5 µm

1875 nm is 1.8 µm