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Schrodinger Equation

The wave func...

Question

The wave function of hydrogen atom with its electron in the 2p state varies with direction as well as distance from the nucleus. What is the probability of a 2p electron, for which ml=0,existing in xy plane is 0.Why?

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Q. The wave function for

1 s

orbital of the hydrogen atom is given by

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

a_{0} =

Radius of first Bohr orbit

r =

Distance from the nucleus (Probability of finding the electron varies with respect to it)

What will be the ratio of probabilities of finding the electrons at the nucleus to first Bohr's orbit

a_{0}

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

1 s

orbital of hydrogen, the maximum probability of finding an electron is at

r =

:

Q. For an electron in a hydrogen atom, the wave function

Ψ

is proportional to exp, where

a_{0}

is the Bohrs radius. What is the ratio of the probability of finding the electron at the nucleus to the probability of finding it at

a_{0}

?

Q. STATEMENT - 1 : In the duration electron jumps from first excited state to ground state in a stationary isolated hydrogen atom, the angular momentum of the electron about the nucleus is conserved.

STATEMENT - 2 : As the electron jumps from first excited state to ground state, in a hydrogen atom, the electrostatic force on the electron is always directed towards the nucleus.

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

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