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Can somebody give me a clue on how to solve the question in the picture for me given below

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closed as off-topic by Avnish Kabaj, NotEvans., Gaurang Tandon, Mithoron, Tyberius Jun 16 '18 at 16:20

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  • $\begingroup$ What do you want us to do with this? $\endgroup$ – Ivan Neretin Jun 16 '18 at 5:41
  • $\begingroup$ Whyever not do it, it might be for the simple joy of doing it. $\endgroup$ – Nuclear Chemist Jun 16 '18 at 9:14
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The normalised wave function of $2s$ orbital is $$\Psi_{2s} = \frac{1}{4 \sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(2 - \frac{Zr}{a_0}\right)e^{-\frac{Zr}{2a_0}}$$ Now the radial probability distribution function will be $$P(r) = |\Psi|^2 4\pi r^2 = \frac{Z^3}{8a_0^3}r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}},$$ so that $\int_{0}^{\infty} P(r) ~\mathrm dr =1.$ Now for getting the maxima and minima, solve $\frac{\mathrm d}{\mathrm dr}(P(r)) = 0.$ Simply we have to solve $$\frac{\mathrm d}{\mathrm dr}\left[r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}}\right] =0.$$ Solving this one root will be from the equation $(2- \frac{Zr}{a_0})=0$, which will give you the value of $ r$ for which $P(r) =0$, that will be the minima as you can easily verify. The maximas will come from the other quadratic equation $$\left(\frac{Zr}{a_0}\right)^2 -6\left(\frac{Zr}{a_0}\right) + 4 =0$$ which will give you two root as $\frac{Zr}{a_0} = 3+ \sqrt{5} ;\, 3-\sqrt{5}.$

So, the difference between two peak points will be $r_d =\frac{2\sqrt{5} a_0}{Z}.$ If it is Hydrogen atom, the distance will be $2\sqrt{5} \times0.529 A^0 = 2.365 A^0.$

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