What is the energy of $He^+$ electron in the first orbit in $eV$?

  • A
    $40.8$
  • B
    $-27.2$
  • C
    $-54.4$
  • D
    $-13.6$

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An electron in a hydrogen atom,after absorbing energy photons,can jump between energy states $n_1$ and $n_2$ $(n_2 > n_1)$. It then returns to the ground state,emitting six different wavelengths in the emission spectrum. The energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. Determine $n_1$ and $n_2$.

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Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is . . . . . . eV.

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The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

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