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 :

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 :

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 $$E = {{1240\,eV} \over {\lambda (in\,nm)}}$$) :

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)