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Citrus2011 [14]
3 years ago
14

In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

electron falls from the fourth to the second orbital, what from the A. 3.04 x 10 Hz (or cycles per second) B. 608 x 1014 ã C. 3.04 x 1015 Hz D. 4.00 x 10'"Hz E. 4.00 x 10 Hz
Physics
1 answer:
agasfer [191]3 years ago
4 0

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

Explanation:

Given that,

Energy E=4.07\times10^{-19}\ J

We need to calculate the energy

Using relation of energy

E_{4}-E_{2}=\Delta E

Where, \Delta E =  energy spacing

4h\nu-2h\nu=4.07\times10^{-19}

\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

\nu=3.069\times10^{14}\ Hz

Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

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