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Expert-verified Found in: Page 1216 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # ${\mathbf{6}}{\mathbf{.}}{\mathbf{2}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{14}}}{\mathbf{}}{\mathbf{Hz}}$An 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 $4.12×{10}^{-19}\mathrm{J}$.

See the step by step solution

## Step 1: The energy of the photon:

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

${\mathbf{∆}}{\mathbf{E}}{\mathbf{=}}{\mathbf{hf}}$ ….. (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 $6.626×{10}^{-34}\mathrm{J}.\mathrm{s}$ for h, and $6.2×{10}^{14}\mathrm{Hz}$ for f in equation (1).

$∆\mathrm{E}=\left(6.626×{10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(6.2×{10}^{14}\mathrm{Hz}\right)\phantom{\rule{0ex}{0ex}}=4.12×{10}^{-19}\mathrm{J}$

Hence, the energy of the atom is increased by $4.12×{10}^{-19}\mathrm{J}$. ### Want to see more solutions like these? 