A) \[\frac{{{N}_{2}}}{{{N}_{1}}}=1.1577\times {{10}^{-38}}\]
B) \[{{E}_{n}}=\frac{-13.6}{({{n}^{2}})}eV\]
C) \[\therefore \]
D) \[E=\frac{-13.6}{{{(5)}^{2}}}eV=-0.54eV\]
Correct Answer: A
Solution :
We have,\[U=k\left[ \frac{2q(8d)}{r}-\frac{(2q)(q)}{x}-\frac{(8q)(q)}{r-x} \right]\] On differentiating both sides w.r.t. x, we get \[k=\frac{1}{4\pi {{\varepsilon }_{0}}}\] \[\frac{2}{x}+\frac{8}{r-x}\]\[\frac{2}{x}+\frac{8}{r-x}=y\] At points of maximum, we must have \[\frac{dy}{dx}=0\] Now,\[-\frac{2}{{{x}^{2}}}+\frac{8}{{{(r-x)}^{2}}}=0\] \[\Rightarrow \]\[\frac{x}{r-x}=\sqrt{\frac{2}{8}}=\frac{1}{2}\Rightarrow x=\frac{r}{3}\] So, y is maximum at x = 1. Clearly, y = 1 for x = 1. Thus, the point of maximum is (1,1). The equation of the tangent at (1,1) is \[x=\frac{r}{3},\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]\[x=\frac{r}{3},y\]\[\frac{{{N}_{2}}}{{{N}_{1}}}\]
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