Answer
- 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
$\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.844\times10^{-19}}\approx4.101\times10^{-7}m\approx410.1nm$