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. An excited hydrogen atom with an electron in the n 5 5 state emits light having a frequency of 6.90 3 1014 s21. Determine the principal quantum level for the final state in this electronic transition.

Respuesta :

pstnonsonjoku

Answer:

2

Explanation:

Given that the energy of a photon of light is obtained from;

ΔE = hf = 6.6 * 10^-34 Js * 6.903 * 10^14 = 45.6 * 10^-20 J

For moving from n=2 -n-=5 the electron absorbed energy hence the energy is positive.

but;

ΔE = -2.18 * 10^-18 (1/n^2final - 1/n^2initial)

45.6 * 10^-20 J= 2.18 * 10^-18 J (1/n^2final - 1/5^2)

0.21 =  (1/n^2final - 1/5^2)

1/n^2final = 0.21 + 1/5^2

n^2final = 4

nfinal = 2

tutorAnne

The principal quantum level for the final state in this electronic transition is 2.

We first find the wavelength of the hydrogen atom Using the formula;

λ = c/f

λ= wavelength  of the photon

h = Plank's constant

f = frequency

λ =3 × 10^8 m/s / 6.903 x 10^14 s-1

λ = 4.3459 × 10^-7 m

Now that we know that

  • Initial  transition = 2
  • Wavelength =4.3459 × 10^-7 m

We can now use the Rydberg equation 1/λ =R(1/n²f -1/n²i)to  determine the principal quantum level in which the electron stopped.

Remember that Rydberg constant =   1.09 x 10^7 m⁻1

1/λ =R(1/n²f -1/n²i)

1/4.3459 × 10^-7 m = 1.09 x 10^7 m⁻1(1/n²f -1/5²)

1/(4.3459 × 10^-7 m x1.09 x 10^7) = (1/n²f -1/5²)

1/4.7370= (1/n²f -1/5²)

0.211= (1/n²f -1/5²)

1/n²f= 0.211+ (1/5²)

1/n²f= 0.211+0.04

1/n²f=0.25

n²f=1/0.25

nf  =   √4

nf =  2

Therefore final transistion is 2.0

Learn more on Calclations involving principal quantum level here:https://brainly.com/question/14651828

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