6. The diagram below shows an emission transition in the Bohr model of the hydro

Question

6. The diagram below shows an emission transition in the Bohr model of the hydrogen atom from n = 3 to n = 2. n=3 n=2 +Ze This is the lowest energy emission transition that corresponds to a line in the Balmer series that we observed during lab. Consider the following equation for the energy of a photon he photon where h is Planck's constant (6.626 × 10-34), c is the speed of light (2.9979 × 108), and is the wavelength. a) The equation above implies that there is an inversely proportional relationship between the energy and the wavelength of emission. There are four main transitions in the Balmer series at 656.3, 486,1, 434.0, and 410.2 nm. which wavelength corresponds to the n 3 to n 2 transition? b) Use the equation above and the wavelength you chose in part (a) in order to calculate the energy of the photon emitted during this transition.

Explanation / Answer

a) As energy is inversely proportional to wavelength

The transition n=3 to n=2 is the lowest energy transition of Balmer series.

Thus it should have the largest wavelength.

Thus the wavelength corresponding to n=3 to n=2 transition is 656.3 nm

b)

Energy corresponding to te n=3 to n=2 transition is

E = hc/ lambda

= 6.626x10-34 J.sec x 2.9979x108 m/sec / 656.3x10-9 m

= 3.027x10-19 J

The energy emitted during the transition = 3.027x10-19 J

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