For 1s orbital of hydrogen atom, the radial electron probability has a maximum value from its nucleus, at a distance of:

0.106 A^{0}

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0.53 A^{0}

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1.06 A^{0}

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2.06 A^{0}

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Solution

The correct option is B $0.53 A^{0}$
The radial electron probability is maximum on the orbit. As its the 1s orbital, therefore, its the first shell. So the radius of the first orbit is $= 0.0529 n m$ $= 0.53 A^{0}$

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Similar questions

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 =

Q. For

1 s

orbital of Hydrogen atom radial wave function is given as:

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

where

a_{0} = 0.529 A \circ)

The ratio of radial probability density of finding electron at

r = a_{o}

to the radial probability density of finding electron at the nucleus is given as (x.e

^{- y}

). Calculate the value of (x + y)

For the $1 s$ orbital of the Hydrogen atom, the radial wave function is given as:
$R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}$ (Where $a_{0} = 0.529 \circ A$ )

The ratio of radial probability density of finding an electron at $r = a_{0}$ to the radial probability density of finding an electron at the nucleus is given as $(x . e^{- y})$ .
Calculate the value of $(x + y)$ .