An ionic atom equivalent to a hydrogen atom has a wavelength equal to $1/4$ of the wavelength of the corresponding hydrogen line. The ion is:

Four electrons,each of mass $m_{e}$,are in a one-dimensional box of size $L$. Assume that the electrons are non-interacting,obey the Pauli exclusion principle,and are described by standing de Broglie waves confined within the box. Define $\alpha = h^{2} / 8 m_{e} L^{2}$ and $U_{0}$ to be the ground state energy. Then,

Imagine removing one electron from $He^4$ and $He^3$. Their energy levels,as worked out on the basis of the Bohr model,will be very close. Explain why?

$A$ particle of mass $m$ moves in a circular orbit in a central potential field $U(r) = \frac{1}{2}kr^2$. If Bohr's quantization conditions are applied,radii of possible orbits and energy levels vary with quantum number $n$ as

An electron in the $n = 1$ orbit of a hydrogen atom is bound by $13.6\, eV$. The energy required to ionize it is........$ eV$.

The figure below is the plot of potential energy versus internuclear distance $(d)$ of $H_2$ molecule in the electronic ground state. What is the value of the net potential energy $E_0$ (as indicated in the figure) in $kJ \ mol^{-1}$, for $d=d_0$ at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of $H$ atom is taken as zero when its electron and the nucleus are infinitely far apart.
Use Avogadro constant as $6.023 \times 10^{23} \ mol^{-1}$.