The radial distribution function P(r) is used to determine the most probable radius, which is used to find the electron in a given orbital. dpdr(r) for ‘1s’ orbital of hydrogen like atom having atomic number z is dpdr=4z3ao3[2r−2zr2ao]e−2zrao, then which of the following statements is/are correct.

At the point of maximum value of radial distribution function d p d r ( r ) = 0 , one antinode is present , Most probable radius of L i 2 + is ( a o 3 ) , Most probable radius of hydrogen atom is ‘ a o ’ , Most probable radius of H e + is a o 2

The radial distribution function P(r) is used to determine the most probable radius, which is used to find the electron in a given orbital. dpdr(r) for ‘1s’ orbital of hydrogen like atom having atomic number z is dpdr=4z3ao3[2r−2zr2ao]e−2zrao, then which of the following statements is/are correct.

When d p d r = 0 , ( 2 r − 2 z r 2 a o ) = 0 r = a o 2

At the point of maximum value of radial distribution function d p d r ( r ) = 0 , one antinode is present

Most probable radius of L i 2 + is ( a o 3 )

Most probable radius of H e + is a o 2

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The radial distribution function P(r) is used to determine the most probable radius, which is used to find the electron in a given orbital. $\frac{d p}{d r} (r)$ for ‘1s’ orbital of hydrogen like atom having atomic number z is $\frac{d p}{d r} = \frac{4 z^{3}}{a_{o}^{3}} [2 r - \frac{2 z r^{2}}{a_{o}}] e^{\frac{- 2 z r}{a_{o}}}, then$ which of the following statements is/are correct.

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At the point of maximum value of radial distribution function $\frac{d p}{d r} (r) = 0,$ one antinode is present

Most probable radius of $L i^{2 +} is (\frac{a_{o}}{3})$

Most probable radius of hydrogen atom is $‘ a_{o} ’$

Most probable radius of $H e^{+} is \frac{a_{o}}{2}$

answer is A, B, C, D.

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Detailed Solution

When $\frac{d p}{d r} = 0, (2 r - \frac{2 z r^{2}}{a_{o}}) = 0$
$r = \frac{a_{o}}{2}$

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