P is the probability of finding the 1s electron of the hydrogen atom...

2 months ago 1

P

$P$ is the probability of finding the

1 s

$1 s$ electron of the hydrogen atom in a spherical shell of infinitesimal thickness,

dr

$dr$ , at a distance,

r

$r$ , from the nucleus. The volume of this shell is

4 π r^{2} d r

$4 π r^{2} d r$ . The qualitative sketch of the dependence of

P

$P$ on

r

$r$ is:

Solution:

For

1 s

$1 s$ , the radial part of the wave function is:

Ψ_{(r)} = 2 {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{- \frac{r}{a_{0}}}

$Ψ_{(r)} = 2 {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{- \frac{r}{a_{0}}}$
Probability of finding an

e^{-}

$e^{-}$ in a spherical shell of thickness,

' dr'

$' dr'$ at a distance

' r'

$' r'$ from the nucleus:

P = Ψ_{(r)}^{2} . 4 π r^{2} d r

$P = Ψ_{(r)}^{2} . 4 π r^{2} d r$

= 16 π r^{2} {(\frac{1}{a_{0}})}^{3} e^{\frac{- 2 r}{a_{0}}} d r

$= 16 π r^{2} {(\frac{1}{a_{0}})}^{3} e^{\frac{- 2 r}{a_{0}}} d r$
So,

P

$P$ is zero at

r = 0

$r = 0$ and

r = \infty

$r = \infty$ .

So, the plot of radial probability function,

4 π r^{2} d r . Ψ_{(r)}^{2}

$4 π r^{2} d r . Ψ_{(r)}^{2}$ V/s

r

$r$ is as shown above.

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P is the probability of finding the 1s electron of the hydrogen atom...

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P is the probability of finding the 1s electron of the hydrogen atom...

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