The wave function -n-l-m is a mathematical function whose value depends upon spherical polar coordinates r- - - of the electron and characterized by the quantum numbers n- l and ml- Here r is distance from nucleus- - is colatitude and - is azimuth- In the mathematical functions given in the table- Z is atomic number and a0 is Bohr radius-For hydrogen atom- the only CORRECT combination is

For H atom 1s orbital ψ_nlm∝ (Z/a_0)^3/2 e^ (zr/a_0) E_4 E_2 = 13.6/16 ( 13.6/4) = 3 × 13.6/16 E_6 E_2 = 13.6/36 ( 13.6/4) = 8 × 13.6/36 E_4 ...

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Byju's Answer

Standard XIII

Chemistry

Introduction to Quantum Mechanical Model

The wave func...

Question

The wave function $ψ_{n, l, m}$ is a mathematical function whose value depends upon spherical polar coordinates $(r, θ, ϕ)$ of the electron and characterized by the quantum numbers n, l and $m_{l}$ . Here r is distance from nucleus, $θ$ is colatitude and $ϕ$ is azimuth. In the mathematical functions given in the table, Z is atomic number and $a_{0}$ is Bohr radius.

For hydrogen atom, the only CORRECT combination is

(I) (iv) (R)

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(I) (i) (P)

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(II) (i) (Q)

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(I) (i) (S)

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Solution

The correct option is D (I) (i) (S)
For H-atom
$1 s o r b i t a l - ψ_{n l m} \propto (\frac{Z}{a_{0}})^{\frac{3}{2}} e^{- (\frac{z r}{a_{0}})}$

$E_{4} - E_{2} = - \frac{13.6}{16} - (- \frac{13.6}{4}) = \frac{3 \times 13.6}{16}$
$E_{6} - E_{2} = - \frac{13.6}{36} - (- \frac{13.6}{4}) = \frac{8 \times 13.6}{36}$
$E_{4} - E_{2}$ is $\frac{27}{32}$ times of $E_{6} - E_{2}$

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Similar questions

Q. The wave function

ψ_{n, l, m}

is a mathematical function whose value depends upon spherical polar coordinates

(r, θ, ϕ)

of the electron and characterized by the quantum numbers n, l and

m_{l}

. Here r is distance from nucleus,

θ

is colatitude and

ϕ

is azimuth. In the mathematical functions given in the table, Z is atomic number and

a_{0}

is Bohr radius.

For hydrogen atom, the only CORRECT combination is

Q. The wave function

ψ_{n, l, m}

is a mathematical function whose value depends upon spherical polar coordinates

(r, θ, ϕ)

of the electron and characterized by the quantum numbers n, l and

m_{l}

. Here r is distance from nucleus,

θ

is colatitude and

ϕ

is azimuth. In the mathematical functions given in the table, Z is atomic number and

a_{0}

is Bohr radius.

For

H e^{+}

ion, the only INCORRECT combination is

The wave function, $Ψ_{n, l, m_{l}}$ is a methematical function whose value depends upon spherical polar coordinates (r, θ, Ф) of the electron and characterized by the quantum numbers n, l and m. Here r is distance from nucleus, θ is colatitude and Ф is azimuth. In the mathematical functions given in the Table, Z is atomic number and $a_{0}$ is Bohr radius.

Column 1	Column 2	Column 3
(i) 1s orbital	(i) $Ψ_{n, l, m_{l}} \propto {(\frac{Z}{a_{0}})}^{\frac{3}{2}} e^{(\frac{- Z r}{a_{0}})}$	(P)
(ii) 2s orbital	(ii) One radial node	(Q) Probability density at nucleus $\propto \frac{1}{a_{0}^{3}}$
(iii) 2 $p_{z}$ orbital	(iii) $Ψ_{n, l, m_{l}} \propto {(\frac{Z}{a_{0}})}^{\frac{5}{2}} r e^{(\frac{- Z r}{2 a_{0}})} c o s θ$	(R) Probability density is maximum at nucleus
(iv) 3 $d_{z}^{2}$ orbital	(iv) xy-plane is a nodal plane	(S) Energy needed to excite electron from n = 2 state to n = 4 state is $\frac{27}{32}$ times the energy needed to excite electron from n = 2 state to n = 6 state

For $H e^{+}$ ion, the only INCORRECT combination is:

Column 1	Column 2	Column 3
(i) 1s orbital	(i) $Ψ_{n, l, m_{l}} \propto {(\frac{Z}{a_{0}})}^{\frac{3}{2}} e^{(\frac{- Z r}{a_{0}})}$	(P)
(ii) 2s orbital	(ii) One radial node	(Q) Probability density at nucleus $\propto \frac{1}{a_{0}^{3}}$
(iii) 2 $p_{z}$ orbital	(iii) $Ψ_{n, l, m_{l}} \propto {(\frac{Z}{a_{0}})}^{\frac{5}{2}} r e^{(\frac{- Z r}{2 a_{0}})} c o s θ$	(R) Probability density is maximum at nucleus
(iv) 3 $d_{z}^{2}$ orbital	(iv) xy-plane is a nodal plane	(S) Energy needed to excite electron from n = 2 state to n = 4 state is $\frac{27}{32}$ times the energy needed to excite electron from n = 2 state to n = 6 state

For $H e^{+}$ ion, the only INCORRECT combination is:

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 =

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The correct option is D (I) (i) (S)For H-atom 1s orbital−ψnlm∝(Za0)32 e−(zra0) E4−E2=−13.616−(−13.64)=3×13.616 E6−E2=−13.636−(−13.64)=8×13.636 E4−E2 is 2732 times of E6−E2