- 1s - 1 - a o 3 - e -r- a 0Above relation is wave function of hydrogen atom in ground state r- distance of electron from nucleus and a 0 -0-529- 10 -8 cm- For 1s orbital of hydrogen- the maximum probability of finding an electron is at r-

The correct option is D a _ 0 /2 Given, Ψ = [1√(π a^3_0)]e^ r / a_0 For maximum stability, #160; Ψ^2 = 0 Ψ^2 = 1π a^3_0e^ r / a_0 ⇒2r ...

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Heisenberg's Uncertainty Principle

Ψ 1s =[ 1 √π ...

Question

$Ψ_{1 s} = [\frac{1}{\sqrt{π a_{o}^{3}}}] e^{- r / a_{0}}$
Above relation is wave function of hydrogen atom in ground state ( $r =$ distance of electron from nucleus and $a_{0} = 0.529 \times 10^{- 8} c m$ ). For $1 s$ orbital of hydrogen, the maximum probability of finding an electron is at $r =$ :

a_{0} / 2

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a_{0}

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2 a_{0}

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\sqrt{a_{0}}

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Solution

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1 s

Ψ_{1 s} = \frac{π}{\sqrt{2}} e^{- r / a_{0}}

a_{0} =

r =

a_{0}

1 s

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

a_{0} = 0.529 \circ A

r = a_{0}

(x . e^{- y})

(x + y)

1 s

R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} e^{- r / a_{0}}

a_{0} = 0.529 A \circ)

r = a_{o}

^{- y}

Ψ = \frac{1}{\sqrt{π}} {[\frac{1}{a_{0}}]}^{3 / 2} e^{- r / a_{0}}

a_{0}

ψ = \frac{1}{4 a_{0} \sqrt{2 π a_{0}}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{2 a_{0}}}

a_{0}

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