Question 3 : Calculate the wavelength in angstroms of electromagnetic radiation emitted by a hydrogen atom which undergoes a transition between energy levels of -1.36×10−19J and ‘-5.45 ×10−19J. (Take Planck??s constant h = 6.6 ×10′−34Js)

Question 3 : Calculate the wavelength in angstroms of electromagnetic radiation emitted by a hydrogen atom which undergoes a transition between energy levels of -1.36×10⁻¹⁹J and ‘-5.45 ×10⁻¹⁹J. (Take Planck??s constant h = 6.6 ×10^′−34Js)
E1– E2 = hf = hc/λ

E1 =  -1.36×10⁻¹⁹J

E2 = 5.45 ×10⁻¹⁹J

6. 6.6 ×10^′−34Js

λ = ?

Solution

E1– E2 = hf = hc/λ

E don’t have frequency so we going to solve for frequency first

Frequency can be calculated using the following equation

E2 – E1 = hf

F = E2 – E1

h

-5.45 ×10⁻¹⁹J + 1.36×10⁻¹⁹J  = 6.6 ×10^′−34JsF making F the subject of the formula we have

F = -5.45 ×10⁻¹⁹J + 1.36×10⁻¹⁹J  = -619696969696969.69696969696969697 approximately is  

                6.6 ×10^′−34Js

= 6.2 X 10¹⁵Hz

F = 6.2 X 10¹⁵Hz

Now we want to look for wavelength

Which is E1– E2 = hc/λ

Λ  =  hc/ E1– E2

 

So we

E1 = -1.8eV

E2 = -4.0eV

-1.8 + 4.0 = 2.2ev

H = 6.6 ×10^′−34Js

C = 3 .0 ×10^′8ms-1

 

6.6 ×10^′−34Js x 3 .0 ×10^′8ms-1    = 0.000000000000000000000000198

Λ =  —————————————————————————————————–   =

2.2 x 3 .0 ×10^′8ms-1             = 0.000000000000000352

=0.0000000005625 approximately = 5.625 x 3 .0 ×10^′-10m  answer

 

 

 

 

 

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