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Atoms and Nuclei question

2023 · 6 Apr · Shift 2 · Q64
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Atoms and Nuclei question

2023 · 6 Apr · Shift 2 · Q64

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
Experimentally it is found that 12.8 eV12.8 ~\mathrm{eV}12.8 eV 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×10−10 m\frac{9}{x} \times 10^{-10} \mathrm{~m}x9​×10−10 m. The value of the xxx is ‾\underline{\hspace{2cm}}​. (1eV=1.6×10−19 J,14πϵ0=9×109Nm2/C2\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}, \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right.(1eV=1.6×10−19 J,4πϵ0​1​=9×109Nm2/C2 and electronic charge =1.6×10−19C)\left.=1.6 \times 10^{-19} \mathrm{C}\right)=1.6×10−19C)
Numerical answer
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Correct answer: 16

  1. Given

    The energy required to separate the hydrogen atom into proton and electron is the ionization energy: E=12.8 eVE = 12.8\ \text{eV}E=12.8 eV

    Converting into joules: E=12.8×1.6×10−19=20.48×10−19=2.048×10−18 JE = 12.8 \times 1.6\times 10^{-19} = 20.48\times 10^{-19} = 2.048\times 10^{-18}\ \text{J}E=12.8×1.6×10−19=20.48×10−19=2.048×10−18 J

  2. Energy relation for electron in hydrogen atom

    For an electron in circular orbit under electrostatic attraction:

    • Potential energy: U=−14πϵ0e2rU = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}U=−4πϵ0​1​re2​
    • Kinetic energy: K=1214πϵ0e2rK = \frac{1}{2}\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}K=21​4πϵ0​1​re2​

    Hence total energy is Etotal=K+U=−1214πϵ0e2rE_{\text{total}} = K+U = -\frac{1}{2}\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}Etotal​=K+U=−21​4πϵ0​1​re2​

    Therefore, the energy required to remove the electron is ∣Etotal∣=1214πϵ0e2r|E_{\text{total}}| = \frac{1}{2}\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}∣Etotal​∣=21​4πϵ0​1​re2​

  3. Substitute given values

    2.048×10−18=12×9×109×(1.6×10−19)2r2.048\times 10^{-18} = \frac{1}{2}\times 9\times 10^9 \times \frac{(1.6\times 10^{-19})^2}{r}2.048×10−18=21​×9×109×r(1.6×10−19)2​

    First compute: (1.6×10−19)2=2.56×10−38(1.6\times 10^{-19})^2 = 2.56\times 10^{-38}(1.6×10−19)2=2.56×10−38

    So, 2.048×10−18=12×9×109×2.56×10−38r2.048\times 10^{-18} = \frac{1}{2}\times 9\times 10^9 \times \frac{2.56\times 10^{-38}}{r}2.048×10−18=21​×9×109×r2.56×10−38​

    2.048×10−18=11.52×10−29r2.048\times 10^{-18} = \frac{11.52\times 10^{-29}}{r}2.048×10−18=r11.52×10−29​

  4. Solve for rrr

    r=11.52×10−292.048×10−18r = \frac{11.52\times 10^{-29}}{2.048\times 10^{-18}}r=2.048×10−1811.52×10−29​

    r=(11.522.048)×10−11r = \left(\frac{11.52}{2.048}\right)\times 10^{-11}r=(2.04811.52​)×10−11

    r=5.625×10−11 mr = 5.625\times 10^{-11}\ \text{m}r=5.625×10−11 m

  5. Compare with required form

    We need r=9x×10−10 mr = \frac{9}{x}\times 10^{-10}\ \text{m}r=x9​×10−10 m

    Also, 5.625×10−11=0.5625×10−105.625\times 10^{-11} = 0.5625\times 10^{-10}5.625×10−11=0.5625×10−10

    Hence, 9x=0.5625\frac{9}{x} = 0.5625x9​=0.5625

    x=90.5625=16x = \frac{9}{0.5625} = 16x=0.56259​=16

  6. Final answer

    16\boxed{16}16​

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