Question

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1. How much energy is in a beta-Lyman emission of hydrogen?

2. The gamma Lyman emission of hydrogen corresponds to an electron dropping from which energy level to which other energy level?

Answer 1

(1)Beta-Lyman emission occurs when the electron falls from the n = 3 orbital to the n = 1 orbital

From Rydburg's Equation , 1/ λ = R[(1/n_i²) – (1/n_f²)]
   Where R = Rydburg's constant = 10.96 x10⁶ m^-1
               λ = wavelength = ?
               n_i = 1
               n_f = 3
Plug the values we get λ = 1.026x10^-7 m

Energy ,E = hc / λ

Where

h = plank's constant = 6.625x 10^-34 Js

c = speed of light = 3x 10⁸ m/s

λ = wave length = 1.026x10^-7 m

Plug the values we get E = 1.94x10^-18 J

(2)Gamma-Lyman emission occurs when the electron falls from the n = 4 orbital to the n = 1 orbital

From Rydburg's Equation , 1/ λ = R[(1/n_i²) – (1/n_f²)]
   Where R = Rydburg's constant = 10.96 x10⁶ m^-1
               λ = wavelength = ?
               n_i = 1
               n_f = 4
Plug the values we get λ = 9.732x10^-8 m

Energy ,E = hc / λ

Where

h = plank's constant = 6.625x 10^-34 Js

c = speed of light = 3x 10⁸ m/s

λ = wave length = 9.732x10^-8 m

Plug the values we get E = 2.042x10^-18 J