calculate the wave length of radiation emitted when an dectra in a hydrogen atom makes a transition from an energy level with n=3 to n=2.

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Solution

In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using:

E = (13.6 eV) [1/n₂² - 1/n₃²]
= -1.13 eV
= -1.13 x 1.6 x 10^-19J = 1.808 x 10^-19J

E = hν

1.808 x 10^-19J = (6.626 x 10¯³⁴ J s) (ν)
ν = 0.273 x 10¹⁵ s^-1

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calculate the wave length of radiation emitted when an dectra in a hydrogen atom makes a transition from an energy level with n=3 to n=2.

In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using: E = (13.6 eV) [1/n22 - 1/n32] = -1.13 eV = -1.13 x 1.6 x 10-19 J = 1.808 x 10-19 J E = hν 1.808 x 10-19 J = (6.626 x 10¯34 J s) (ν) ν = 0.273 x 1015 s-1

In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using:

E = (13.6 eV) [1/n₂² - 1/n₃²]
= -1.13 eV
= -1.13 x 1.6 x 10^-19J = 1.808 x 10^-19J

E = hν

1.808 x 10^-19J = (6.626 x 10¯³⁴ J s) (ν)
ν = 0.273 x 10¹⁵ s^-1