Question

Consider the electronic transition from n = 4 to n = 1 in a hydrogen atom, and select the correct statement below: A photon of 97 nm wavelength and 2.05x10-18 J energy was emitted from the hydrogen atom in this electronic transition. A photon of 97 nm wavelength and 2.05x10-18 J energy was absorbed by the hydrogen atom in this electronic transition. A photon of 122 nm wavelength and 1.64x10-18 J energy was emitted from the hydrogen atom in this electronic transition. A photon of 122 nm wavelength and 1.64x10-18 J energy was absorbed by the hydrogen atom in this electronic transition.

Answer #1

The correct statement is the first one

A photon of 97 nm wavelength and 2.05x 10^{-18} J of
energy was emitted during the transition.

Since the electron is jumping from n=4 to n=1 , that is from higher level to lower level , energy will be emitted in the form of radiation.

energy emitted = E_{4} - E_{1} =-
2.18x 10^{-18} /16 -(- 2.18x 10^{-18} )

= +2.04x 10^{-18}J

We know hc/ wavelength = energy and substituting the values of
h=6.623x10^{-34} Jsec and c=3.0x10^{8}m/sec

we get wavelength = 96.8 nm

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