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Q38P

Expert-verifiedFound in: Page 1216

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**${\mathbf{6}}{\mathbf{.}}{\mathbf{2}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{14}}}{\mathbf{}}{\mathbf{Hz}}$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 $4.12\times {10}^{-19}\mathrm{J}$.

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

${\mathbf{\u2206}}{\mathbf{E}}{\mathbf{=}}{\mathbf{hf}}$ ….. (1)

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

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

$\u2206\mathrm{E}=\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(6.2\times {10}^{14}\mathrm{Hz}\right)\phantom{\rule{0ex}{0ex}}=4.12\times {10}^{-19}\mathrm{J}$

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

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