The first orbital of H is represented by ψ=(1/√π)((1/a0))3 / 2 e-π / a0, where a0 is Bohr's radius. The probability of finding the electron at a distance r from the nucleus in the region d V is

Question

The first orbital of H is represented by ψ=(1/√π)((1/a0))3 / 2 e-π / a0, where a0 is Bohr's radius. The probability of finding the electron at a distance r from the nucleus in the region d V is (A) ψ2dr  (B) ∫ Ψ2 4 π r2 d V (C) Ψ2 4 π r 2 dr (D) ∫ Ψ dV

Tardigrade Admin · Accepted Answer

This is obtained by the solution of Schrodinger wave equation.
Probability

= ψ^{2} d V

Its orbital is spherically symmetrical. Thus,
volume,

V = \frac{4}{3} π r^{3}

∴ \frac{d V}{d r} = 4 π r^{2}

∵

Probability

= ψ^{2} 4 π r^{2} d r .

Q. The first orbital of $H$ is represented by $ψ = \frac{1}{π} (\frac{1}{a _{0}})^{3/2} e^{- π / a_{0}}$ , where $a_{0}$ is Bohr's radius. The probability of finding the electron at a distance $r$ from the nucleus in the region $d V$ is

$ψ^{2} d r$

$\int Ψ^{2} 4 π r^{2} d V$

$Ψ^{2} 4 π r^{2} d r$

$\int Ψ d V$

This is obtained by the solution of Schrodinger wave equation.
Probability $= ψ^{2} d V$
Its orbital is spherically symmetrical. Thus,
volume, $V = \frac{4}{3} π r^{3}$
$∴ \frac{d V}{d r} = 4 π r^{2}$
$∵$ Probability $= ψ^{2} 4 π r^{2} d r .$

Q. The first orbital of H is represented by ψ=π​1​(a0​1​)3/2e−π/a0​, where a0​ is Bohr's radius. The probability of finding the electron at a distance r from the nucleus in the region dV is

ψ2dr

∫Ψ24πr2dV

Ψ24πr2dr

∫ΨdV