One of the maximum value of R2 lies approximately atThe r-dependent wave function of hydrogen atom is given by the expressionR=( constant )2ra0−r23a02exp⁡r/3a0Answers the following questions.

A
$r = 2.0 a_{0}$
B
$r = 4.0 a_{0}$
C
$r = 11.47 a_{0}$
D
$r = 12.5 a_{0}$

Solution:

For the maximum or minimum, the first derivative of R with respect tor will be equal to zero

$\frac{d R}{d r} = \frac{d}{d r} [\{const \times (3 - \frac{2 r}{a_{0}} + \frac{2 r^{2}}{9 a_{0}^{2}})\} e^{- r / 3 a_{0}}] = [(- \frac{2}{a_{0}} + \frac{4 r}{9 a_{0}^{2}}) + (3 - \frac{2 r}{a_{0}} + \frac{2 r^{2}}{9 a_{0}^{2}}) (- \frac{1}{3 a_{0}})] e^{- r / 3 a_{0}}$

Hence $- \frac{2}{a_{0}} + \frac{4 r}{9 a_{0}^{2}} - \frac{1}{a_{0}} + \frac{2 r}{3 a_{0}^{2}} - \frac{2 r^{2}}{27 a_{0}^{3}} = 0$

or $\frac{2 r^{2}}{27 a_{0}^{2}} - \frac{10 r}{9 a_{0}} + 3 = 0$ or $2 {(\frac{r}{a_{0}})}^{2} - 30 (\frac{r}{a_{0}}) + 81 = 0$

Solving for $r / a_{0}$ , we get

$\frac{r}{a_{0}} = \frac{30 \sqrt{900 - 8 \times 81}}{4} = \frac{30 \sqrt{252}}{4} = \frac{30 15.87}{4} = 3.53, 11.46$

One of the maximum value of $R^{2}$ lies approximately at
The r-dependent wave function of hydrogen atom is given by the expression
$R = (constant) (\frac{2 r}{a_{0}} - \frac{r^{2}}{3 a_{0}^{2}}) \exp (r / 3 a_{0})$
Answers the following questions.

Solution:

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