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Chapter 39: Q38P (page 1216)

6.2脳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

Expert verified

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

Step by step solution

01

The energy of the photon:

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

鈭�E=hf 鈥�.. (1)

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

02

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

Substitute 6.626脳10-34J.sfor h, and 6.2脳1014Hzfor f in equation (1).

鈭�E=6.626脳10-34J.s6.2脳1014Hz=4.12脳10-19J

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

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Most popular questions from this chapter

Calculate the radial probability density P(r) for the hydrogen atom in its ground state at (a) r = 0 , (b) r = a , and (c) r = 2a, where a is the Bohr radius.

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