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Fundamentals Of Physics
Found in: Page 1216

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Short Answer

6.2×1014 HzAn atom (not a hydrogen atom) absorbs a photon whose associated frequency is . By what amount does the energy of the atom increase?

The energy of the atom is increased by 4.12×10-19 J.

See the step by step solution

Step by Step Solution

Step 1: The energy of the photon:

The expression of the energy of the photon is given by,

E=hf ….. (1)

Here, h is plank constant, and f is frequency of the photon.

Step 2: Define the amount of increase in the energy of the atom:

Substitute 6.626×10-34 J.s for h, and 6.2×1014 Hz for f in equation (1).

E=6.626×10-34 J.s6.2×1014 Hz =4.12×10-19 J

Hence, the energy of the atom is increased by 4.12×10-19 J.

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