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Verify that the ground state energy $E_{0}$ is 13.6 ev by using $E_{0}=\frac{2 \pi^{2} q_{e}^{4} m_{e} k^{2}}{h^{2}}$
$\approx 13.6 \mathrm{eV}$
Physics 103
Chapter 30
Atomic Physics
Cornell University
University of Michigan - Ann Arbor
University of Washington
University of Winnipeg
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In this problem we are asked to verify the grams energy is equal to 13.6 com volts. So we are asked to prove this, whereas the grand state energy is given by given by formula which is 2 pi square times the charge on the electron 4 times. The mass of electron times the colum constant square, divided by the h square, plank constant square, so charge an electron. We know here is 1.6 times 10 to the power minus minus 19, where, as the mass of electron is given as 9.11 times 10 to the power minus 31 kg and k is the coulomb constant, which is 9 times 9 times 10 to the power of 9 Times 9 newton meter square meter, square coulomb square, so h here is 6.6 6.63 times 10 to the power minus 3434 joules joule. Second, so substituting these values in above equation gives us the energy in joules, which is a 2.1716 times 10 to the power 10 to the power 18 jules. So we can convert these joules into electron volts by divided by 1.60 times 10 to the power minus 19. Joule so 1 electron mole has got this amount of energy and dividing this we get 13.6 electron volts and of the problem. Thank you.
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