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Anna11 [10]
3 years ago
9

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

is the energy difference between the two energy levels, in J?
Physics
1 answer:
yarga [219]3 years ago
6 0

Answer:

\Delta E=1.64*10^{-18}J

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

\Delta E=h\nu(1)

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

\nu=\frac{c}{\lambda}(2)

Here c is the speed of light. Replacing (2) in (1):

\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J

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