Chapter 39: Q38P (page 1216)

$6.2 \times 10^{14} Hz$ An atom (not a hydrogen atom) absorbs a photon whose associated frequency is . By what amount does the energy of the atom increase?

Short Answer

Expert verified

The energy of the atom is increased by $4.12 \times 10^{- 19} J$ .

Step by step solution

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.

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

Substitute $6.626 \times 10^{- 34} J . s$ for h, and $6.2 \times 10^{14} Hz$ for f in equation (1).

$∆ E = (6.626 \times 10^{- 34} J . s) (6.2 \times 10^{14} Hz) = 4.12 \times 10^{- 19} J$

Hence, the energy of the atom is increased by $4.12 \times 10^{- 19} J$ .

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$6.2 \times 10^{14} Hz$ An atom (not a hydrogen atom) absorbs a photon whose associated frequency is . By what amount does the energy of the atom increase?

Short Answer

Step by step solution

The energy of the photon:

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

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

Short Answer

Step by step solution

The energy of the photon:

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Translational Dynamics

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Study anywhere. Anytime. Across all devices.

$6.2 \times 10^{14} Hz$ An atom (not a hydrogen atom) absorbs a photon whose associated frequency is . By what amount does the energy of the atom increase?