• question_answer At what temperature will the translational kinetic energy of H-atom equal to that for H-atom of first line Lyman transition? (Given ${{N}_{A}}=6\times {{10}^{23}}$) A)  780 K                          B)  $1.32\times {{10}^{5}}\,K$ C)  $7.84\times {{10}^{4}}K$             D)  $1900\,K$

$q'=(Q-q)$ where,      $\therefore$ $F=\frac{1}{r\pi {{\varepsilon }_{0}}}\frac{Q(Q-q)}{{{R}^{2}}}$ $\frac{dF}{dq}=0$ $\frac{d}{dq}\,\left[ \frac{Kq(Q-q)}{{{R}^{2}}} \right]=0$ $\Rightarrow$ $[Q-2q]=0$  $q=\frac{Q}{2}$ of H atom $U=\frac{{{Q}^{2}}}{2C}$ $C'=KC$ $U'=\frac{{{Q}^{2}}}{2C'}=\frac{1}{2}\,\frac{{{Q}^{2}}}{KC}$