The wave function - n - l - m is a mathematical function whose value depends upon spherical polar coordinates r - - - of the electron and characterizedby the quantum numbers n - 1 and m 1- Here r is distance from nucleus- - is colatitude and - is azimuth- In the mathematical functions given in the table-Z is atomic number and a 0 is Bohr radius-or hydrogen atom- the only CORRECT combination isA- I i PB- II i QC- I i SD- I iv R

The correct option is C (I) (i) (S)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. ...

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Byju's Answer

Standard XII

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 the given orbital in Column-I, the only CORRECT combination for any hydrogen-like species 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:

Q. Consider following Schrodinger wave equation for an orbital of hydrogen atom.

(a_{0}

is first Bohr radius)

Ψ = \frac{\sqrt{2 r}}{81 a_{0}^{2} \sqrt{π a_{0}}} (6 - \frac{r}{a_{0}}) e^{\frac{- r}{3 a_{0}}} c o s θ

\begin{matrix} List-I & List-II (I) & Which orbital is this? & (P) & 1 (I I) & Number of total node(s) & (Q) & 2 (I I I) & Nodal plane & (R) & 3 p_{z} (I V) & Number of radial node(s) & (S) & 3 p_{x} (T) & XY plane (U) & YZ plane \end{matrix}

Match the correct combination considering List-I and List-II

Q. Type of terms and corresponding number is written in Column 2 and Column 3 respectively of the Binomial in Column 1.

\begin{matrix} C o l u m n I & C o l u m n I I & C o l u m n 3 (I) & {(5^{\frac{1}{6}} + 7^{\frac{1}{9}})}^{1824} & (i) & T o t a l n u m b e r o f r a t i o n a l t e r m s & (P) & 4 (I I) & {(5^{\frac{1}{6}} + 2^{\frac{1}{8}})}^{100} & (i i) & T o t a l n u m b e r o f i r r a t i o n a l t e r m s & (Q) & 102 (I I I) & {(3^{\frac{1}{4}} + 4^{\frac{1}{3}})}^{99} & (i i i) & \frac{1}{2} [\frac{T o t a l n u m b e r o f t e r m s}{N u m b e r o f r a t i o n a l t e r m s}], & (R) & 224 w h e r e [.] r e p r e s e n t s g r e a t e s t i n t e g e r f u n c t i o n . (I V) & {(7^{\frac{1}{3}} + 11^{\frac{1}{9}})}^{2007} & (i v) & T o t a l n u m b e r o f t e r m s & (S) & I s d i v i s i b l e b y 13 \end{matrix}

Select the correct combination

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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