A free hydrogen atom after absorbing a photon of wavelength $$\lambda $$_a gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength $$\lambda $$_e. Let the change in momentum of atom due to the absorption and the emission be $$\Delta {p_a}$$ and $$\Delta {p_e}$$, respectively. If $${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$$, which of the option(s) is/are correct? [Use hc = 1242 eVnm; 1 nm = 10^-9 m, h and c are Planck's constant and speed of light in vacuum, respectively]

In a radioactive decay chain, $${}_{90}^{232}Th$$ nucleus decays to $${}_{82}^{212}Pb$$ nucleus. Let $${N_\alpha }$$ and $${N_\beta }$$ be the number of $$\alpha $$ and $${\beta ^ - }$$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principle quantum number n, where n >> 1. Which of the following statement(s) is(are) true?

A fission reaction is given by $$_{92}^{236}U \to _{54}^{140}Xe + _{38}^{94}Sr + x + y$$, where x and y are two particles. Considering $$_{92}^{236}U$$ to be at rest, the kinetic energies of the products are denoted by $${K_{Xe}},{K_{Sr}},{K_x}(2MeV)$$ $$ \text { and } \mathrm{K}_{\mathrm{y}}(2 \mathrm{MeV}) $$, respectively. Let the binding energies per nucleon of $$_{92}^{236}U$$, $$_{54}^{140}Xe$$ and $$_{38}^{94}Sr$$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are