# Chapter 6 - Electronic Structure of Atoms - Exercises: 6.39a

- The energy of an electron when n = 2 is $-5.45\times10^{-19}J$. - The energy of an electron when n = 6 is $-6.056\times10^{-20}J$. - The wavelength of the radiation released is $410.1nm$.

#### Work Step by Step

*We would use this following formula in this exercise to calculate the energy of an electron at a state n: $$E=(-h\times c\times R_h)\times(\frac{1}{n^2})=(-2.18\times10^{-18}J)\times(\frac{1}{n^2})$$ *The energy of an electron when n = 2 is: $E_{n=2}=(-h\times c\times R_h)\times(\frac{1}{n^2})=(-2.18\times10^{-18}J)\times(\frac{1}{4})=-5.45\times10^{-19}J$ *The energy of an electron when n = 6 is: $E_{n=6}=(-h\times c\times R_h)\times(\frac{1}{n^2})=(-2.18\times10^{-18}J)\times(\frac{1}{36})\approx-6.056\times10^{-20}J$ *Calculate the wavelength of radiation released when an electron moves from n = 6 to n = 2. - The change of energy as an electron moves from n = 6 to n = 2 is: $\Delta E= E_{n=2}-E_{n=6}= (-5.45\times10^{-19})-(-6.056\times10^{-20})\approx-4.844\times10^{-19}J$ The absolute value of the change of energy would be equal to the energy of a photon released in the electronic transition (since the value of energy cannot be negative). Therefore, $E_{photon}=|\Delta E|\approx4.844\times10^{-19}J$ From here, we can calculate the wavelength of the radiation released, using this equation: $E_{photon}=\frac{h\times c}{\lambda}$ Therefore, we have $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.844\times10^{-19}}\approx4.101\times10^{-7}m\approx410.1nm$

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