As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion

  • A

    kinetic energy decreases, potential energy increases but total energy remains same

  • B

    kinetic energy and total energy decrease but potential energy increases

  • C

    its kinetic energy increases but potential energy and total energy decrease

  • D

    kinetic energy, potential energy and total energy decrease

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A proton is fired from very far away towards a nucleus with charge $Q=120 \ e$, where $e$ is the electronic charge. It makes a closest approach of $10 \ fm$ to the nucleus. The de Brogle wavelength (in units of $fm$ ) of the proton at its start is :

(take the proton mass, $m _0=(5 / 3) \times 10^{-27} kg , h / e =4.2 \times 10^{-15} J / s / C ; \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 m / F ; 1 fm =10^{-15} m$ ) 

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An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze.$ Then the distance of closest approach for the alpha nucleus will be proportional to

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Energy levels $A, B$ and $C$ of a certain atom correspond to increasing values of energy, i.e. $E_A < E_B < E_C$ . If $\lambda _1, \lambda _2$ and $\lambda _3$ are the wavelengths of radiations corresponding to transitions $C$ to $B, B$ to $A$ and $C$ to $A$ respectively , which of the following relations is correct ?