If the electron in a hydrogen atom jumps from an orbit with level $n_1=2$ to an orbit with level $n_2=1$, the emitted radiation has a wavelength given by

Magnetic field at the centre of a hydrogen-like atom (atomic number $= Z$) due to the motion of an electron in the $n^{th}$ orbit is proportional to

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,

For $He^{+}$, a transition takes place from the orbit of radius $105.8 \ pm$ to the orbit of radius $26.45 \ pm$. The wavelength (in $nm$) of the emitted photon during the transition is. . . . .
[Use: Bohr radius, $a_0=52.9 \ pm$; Rydberg constant, $R_H=2.2 \times 10^{-18} \ J$; Planck's constant, $h=6.6 \times 10^{-34} \ J \ s$; Speed of light, $c=3 \times 10^8 \ m \ s^{-1}$]

An excited $He^+$ ion emits two photons in succession, with wavelengths $108.5 \, nm$ and $30.4 \, nm$, in making a transition to the ground state. The quantum number $n$, corresponding to its initial excited state is $n = ........$ (for a photon of wavelength $\lambda$, energy $E = \frac{1240 \, eV}{\lambda \, (in \, nm)}$).

In a hydrogen atom, the electron makes $6.6 \times 10^{15}$ revolutions per second around the nucleus in an orbit of radius $0.5 \times 10^{-10} \, m$. This is equivalent to a current of nearly: