Electrostatics

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Question 20 of 311

Experimentally it is found that 12.8eV12.8eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is 9x×10−10m\dfrac{9}{x}\times {10}^{-10}m.
The value of the xx is: _____.
(1eV=1.6×10−19J,14πϵ0=9×109Nm2C21eV=1.6\times {10}^{-19}J,\dfrac{1}{4\pi {\epsilon }_{0}}=9\times {10}^{9}\dfrac{{Nm}^{2}}{{C}^{2}} and electronic charge =1.6×10−19C=1.6\times {10}^{-19}C)

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Answer: 16

The given data is
q=1.6×10−19C14πϵ0=9×109Nm2C−2q=1.6\times {10}^{-19}C \dfrac{1}{4{\pi \epsilon }_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2}
The formula for the value of energy in a hydrogen atom is given by
E=q28πϵ0r⇒E=q22(4πϵ0)rE=\dfrac{{q}^{2}}{8\pi {\epsilon }_{0}r} \Rightarrow E=\dfrac{{q}^{2}}{2(4\pi {\epsilon }_{0})r}
⇒r=9×109×1.6×10−1912.8×2\Rightarrow r=\dfrac{9\times {10}^{9}\times 1.6\times {10}^{-19}}{12.8\times 2}
⇒r=916×10−10m\Rightarrow r=\dfrac{9}{16}\times {10}^{-10}m
It is given that r=9x×10−10m⇒916×10−10=9x×10−10⇒x=16r=\dfrac{9}{x}\times {10}^{-10}m \Rightarrow \dfrac{9}{16}\times {10}^{-10}=\dfrac{9}{x}\times {10}^{-10} \Rightarrow x=16

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