Quantum Chemistry/Example 19

The energy of the hydrogen atom is calculated using the following equation:

'''Energy of Hydrogen Atom''' ${\displaystyle E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}}$

To calculate the change in energy required to go from the ground state to n=2, the following equation is used:

${\displaystyle \delta E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}+\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{n^{2}}}}$

${\displaystyle \delta E=-\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{4}}+\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {1}{1}}}$

${\displaystyle \delta E=\left({\frac {m_{e}e^{4}}{32\pi ^{2}\epsilon _{0}^{2}h^{2}}}\right){\frac {5}{4}}}$

${\displaystyle \delta E=\left({\frac {(9.10938x10)^{-31}kg(1.60218x10^{-19}c)^{4}}{32\pi ^{2}8.854x10^{1}2c^{2}S^{2}/kgm^{2})^{2}(1.054x10^{-24}J.s)^{2}}}\right){\frac {5}{4}}}$

${\displaystyle \delta E=\left({\frac {(9.10938x10)^{-31}kg(1.60218x10^{-19}c)^{4}}{32\pi ^{2}8.854x10^{1}2c^{2}S^{2}/kgm^{2})^{2}(1.054x10^{-24}J.s)^{2}}}\right){\frac {5}{4}}}$

δE =((9.10938x10)−31kg(1.60218x10−19c)432π28.854x1012c2S2/kgm2)2(1.054x10−24J.s)2)54