The Bohr model gives a simple, but reasonably accurate, formula for the energy levels of an electron in an isolated hydrogen atom. When an electron moves from one energy level to another lower level, the difference in the energies between the two levels is emitted in a photon. What is the wavelength of the photon emitted when an electron falls from the fourth level to the first level?

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kobenhavn kobenhavn · Answer 1 · 2019-07-12T10:25:08+02:00

Answer: [tex]0.98\times 10^{-7}m[/tex]

Explanation:

For calculating wavelength, when the electron will jump from n=4 to n= 1

Using Rydberg's Equation: for hydrogen atom

[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2[/tex]

Where,

[tex]\lambda[/tex] = Wavelength of radiation = ?

[tex]R_H[/tex] = Rydberg's Constant = [tex]1.097\times 1067m[/tex]

[tex]n_f[/tex] = Higher energy level = 4

[tex]n_i[/tex]= Lower energy level = 1

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

[tex]\frac{1}{\lambda}=1.097\times 10^7\left(\frac{1}{1^2}-\frac{1}{4^2} \right )\times 1^2[/tex]

[tex]\lambda=0.98\times 10^{-7}m[/tex]

Thus the wavelength of the photon emitted will be [tex]0.98\times 10^{-7}m[/tex]