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vladimir1956 [14]
3 years ago
8

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

46.3 nm is emitted. The energy difference between these 2p and 2s orbitals is:_________a. 3.07 Ã 10^â28 Jb. 3.07 Ã 10^â19 Jc. 3.07 Ã 10^â17 Jd. 1.28 Ã 10^â31 Je. none of these
Chemistry
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

Explanation:

Wavelength of the photon emitted = \lambda =646.3 nm =646.3\times 10^{-9} m

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

E=\frac{hc}{\lambda }

h = Planck's constant = 6.626\tiomes 10^{-34} Js

c = Speed of the light = 3\times 10^8 m/s

E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}

E=3.07\times 10^{-19} J

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

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