Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 $$\mathop A\limits^o $$). The de-Broglie wavelength of this electron is :

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $${{{\lambda _1}} \over {{\lambda _2}}}$$, of the photons emitted in this process is :

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

Two radioactive substances A and B have decay constants 5$$\lambda $$ and $$\lambda $$ respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $${\left( {{1 \over e}} \right)^2}$$ will be :