Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstance. calculate the ground state energy of a $H e -$ atom

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Solution

For H atom $Z = 1$ and $n = 1$
$∴ E_{n} = \frac{- m_{g} e^{4}}{8 ε_{0}^{2} 1^{2} h^{2}} = 13.6 e V$
For He- atom, $Z = 4$ and $n = 1$
$∴ E_{n} = \frac{- m_{g} e^{4}}{8 ε_{0}^{2} 1^{2} h^{2}} = - 4 \times 13.6 = 54.4 e V$

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Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstance. calculate the ground state energy of a He−atom

For H atom Z=1 and n=1∴En=−mge48ε2012h2=13.6eVFor He- atom, Z=4 and n=1∴En=−mge48ε2012h2=−4×13.6=54.4eV

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstance. calculate the ground state energy of a $H e -$ atom

For H atom $Z = 1$ and $n = 1$
$∴ E_{n} = \frac{- m_{g} e^{4}}{8 ε_{0}^{2} 1^{2} h^{2}} = 13.6 e V$
For He- atom, $Z = 4$ and $n = 1$
$∴ E_{n} = \frac{- m_{g} e^{4}}{8 ε_{0}^{2} 1^{2} h^{2}} = - 4 \times 13.6 = 54.4 e V$