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 :

In Li^{+ +}, electron in first Bohr orbit is excited to a level by a radiation of wavelength $$\lambda $$. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $$\lambda $$?
(Given : H = 6.63 × 10^–34 Js; c = 3 × 10⁸ ms ^–1)

Two radioactive materials A and B have decay constants 10$$\lambda $$ and $$\lambda $$, respectively. It initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time :