In the following atoms and molecules,for the transition from $n = 2$ to $n = 1$,the spectral line of minimum wavelength will be produced by:

In the Bohr model of the $H$-atom,the kinetic energy of an electron in any orbit depends on the principal quantum number $n$ as:

An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is

Speed of an electron in Bohr's $7^{\text{th}}$ orbit for Hydrogen atom is $3.6 \times 10^6\,m/s$. The corresponding speed of the electron in $3^{\text{rd}}$ orbit,in $m/s$ is $........\times 10^6$.

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 $E = \frac{1240 \, eV}{\lambda \, (in \, nm)}$).

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,