# Chapter 6 - Electronic Structure of Atoms - Exercises: 6.32b

The total energy of photons detected in 1 hour is $6.448\times10^{-11}J$.

#### Work Step by Step

*Strategy: 1) Calculate the energy of a photon of this radiation. 2) Find the amount of energy of photons detected per second. 3) Figure out the total energy of photons detected in 1 hour. 1) Calculate the energy of a photon of this radiation. - Wavelength of radiation: $\lambda=3.55\times10^{-3}m$ - Planck's constant: $h\approx6.626\times10^{-34}J.s$ - Speed of light in a vacuum: $c\approx2.998\times10^8m/s$ The energy of a photon of this radiation is: $E_p=\frac{h\times c}{\lambda}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.55\times10^{-3}}\approx5.596\times10^{-23}J$ 2) Find the energy of photons detected per second. - The number of photons detected per second: $N=3.2\times10^8photons/s$ The energy of photons detected per second is: $E_s=E_p\times N= (5.596\times10^{-23})\times(3.2\times10^8)\approx1.791\times10^{-14}J/s$ 3) Figure out the total energy of photons detected in 1 hour. - Amount of time: $t=1h=3600s$ The total energy of photons detected in 1 hour is: $E=E_s\times t=(1.791\times10^{-14})\times3600\approx6.448\times10^{-11}J$

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