1) Calculate the shortest wavelength associated with emitted photons in the Lyma

Question

1) Calculate the shortest wavelength associated with emitted photons in the Lyman series. (Express your answer to four significant figures.)(In nm).

2)Calculate the highest energy associated with emitted photons in the Lyman series. (Express your answer to three significant figures.) (in ×10^18J).

3.Calculate the shortest wavelength associated with emitted photons in the Balmer series. (Express your answer to four significant figures.) (in nm).

4.Calculate the highest energy associated with emitted photons in the Balmer series. (Express your answer to three significant figures.) (in ×10^19J).

5. Calculate the shortest wavelength associated with emitted photons in the Paschen series. (Express your answer to four significant figures.) (in nm).

6.)Calculate the highest energy associated with emitted photons in the Paschen series. (Express your answer to three significant figures.)(In ×1019J).

Explanation / Answer

1) In physics and chemistry, the Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n2 2 to n1 = 1 (where n is the principal quantum number) the lowest energy level of the electron.

Using Rydberg formula

1/ = RH (1/n1^2 - 1/n2^2) RH = 1.097*10^7 m^-1

shortest wavelenght will be associated with n2 = infinity

= 91.15 * 10^-9 m = 91.15 nm

2) E = -13.6/n^2 eV

For highest energy n = 2

E = -13.6/4 = -3.4 eV = - 5.45e-19 Joules

3) The Balmer series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n2 3 to n1 = 2 (where n is the principal quantum number) the lowest energy level of the electron.

Using Rydberg formula

1/ = RH (1/n1^2 - 1/n2^2) RH = 1.097*10^7 m^-1

shortest wavelenght will be associated with n1 = 2; n2 = infinity

= 3.64*10^-7 m = 364 nm

4) For highest energy n = 3

E = -13.6 / 9 = -1.51 eV = 2.42e-19 J

5)

The Paschen series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n2 4 to n1 = 3 (where n is the principal quantum number) the lowest energy level of the electron.

Using Rydberg formula

1/ = RH (1/n1^2 - 1/n2^2) RH = 1.097*10^7 m^-1

shortest wavelenght will be associated with n1 = 3; n2 = infinity

= 8.2*10^-7 m = 820 nm

6)

For highest energy n = 4

E = -13.6 / 16 = -0.85 eV = 1.36e-19 J

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