The energy levels of an atom is shown in figure.

Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?

Given (h = 6.62 $$\times$$ 10$$^{-34}$$ Js)

The ratio of the density of oxygen nucleus ($$_8^{16}O$$) and helium nucleus ($$_2^{4}\mathrm{He}$$) is

A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 $$\times$$ 10$$^{-15}$$ eVs) :

Consider the following radioactive decay process

$$_{84}^{218}A\buildrel \alpha \over \longrightarrow {A_1}\buildrel {{\beta ^ - }} \over \longrightarrow {A_2}\buildrel \gamma \over \longrightarrow {A_3}\buildrel \alpha \over \longrightarrow {A_4}\buildrel {{\beta ^ + }} \over \longrightarrow {A_5}\buildrel \gamma \over \longrightarrow {A_6}$$

The mass number and the atomic number of A$$_6$$ are given by :