Structure of Atom

Under hypothetical situation, the value of l is greater than n which varies from 0 to n + 1.

For n = 1, l = 0, 1, 2

n = 2, l = 0, 1, 2, 3

Elements follow the following electronic configuration

1s 1p 1d 2s 2p 2d 2f

Atomic number (Z) = 9
1s² 1p⁶ 1d¹

Atomic number = 6
1s² 1p⁴

Atomic number 8
1s² 1p⁶

Atomic number 13
1s² 1p⁶ 1d⁵ (half filled)

For n = 4 possible value of l = 0, 1, 2, 3.

Only l = 2 and l = 3 can have m = -2

$ \therefore $ 4d & 4f subshell associated with n = 4, m = –2.

Internal energy of ‘Ar’ or any gas, has nothing to do with Quantum nature of atom.

Photoelectric effect, atomic spectrum and Black body radiations may be explained by quantum theory.

According to Bohr’s model

${r_n} = {{{n^2}} \over Z} \times {a_0}$

Also 2$\pi $${r_n}$ = n$\lambda $

$ \Rightarrow $ 2$\pi $${{{n^2}} \over Z} \times {a_0}$ = n$\lambda $

$ \Rightarrow $ $\lambda $ = $2\pi \times {n \over Z} \times {a_0}$

For n = 4 and Z = 1

$ \Rightarrow $ $\lambda $ = 8$\pi $${a_0}$

${r_n} = {{{n^2}{a_0}} \over Z}$

For 2^nd Bohr orbit of Li⁺²

n = 2

and Z = 3

r = ${{{2^2}{a_0}} \over 3}$ = ${{4{a_0}} \over 3}$

Hydrogen has three isotopes

(A) Protium (${}_1^1H$) has 0 neutron.

(B) Deutrium (${}_1^2H$) has 1 neutrons.

(C) Tritium (${}_1^3H$) has 2 neutrons.

Total number of neutrons in three
isotopes of hydrogen = 0 + 1 + 2 = 3

For balmer series : n₁ = 2, n₂ = 3, 4, 5, .....$\infty $

For longest wavelength n₂ = 3

${1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$

As wavelength decreases the lines in the Balmer series converge. The correct statements are (I), (II) and (III).

Total number of orbitals
= n² = (5)² = 25

Probability of finding an electron is given by 4$\pi $r²dr ${\psi ^2}$ and at both 'a' and 'c' it will have maximum value.

P[15] = [Ne] 3s²3p³

Group number = 10 + valance shell electron = 10 + 5 = 15

number of valence electrons = 5

Valency = 3

Shortest wavelength can found when n₂ = $\infty $

${\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}$

Here n₁ = series number.

$ \Rightarrow $${\lambda _{shortest}} =$ ${{{R{Z^2}} \over {n_1^2}}}$ = ${{{R{{\left( 1 \right)}^2}} \over {n_1^2}}}$

$ \therefore $ ${{{\lambda _A}} \over {{\lambda _B}}} = {\left( {{{{n_B}} \over {{n_A}}}} \right)^2}$

Given, ${\left( {{{{n_B}} \over {{n_A}}}} \right)^2} = 9$

$ \Rightarrow $ ${{{n_B}} \over {{n_A}}} = 3$

$ \therefore $ Series number of B = 3 $ \times $ Series number of A

So,

If series number of A is Lyman( n = 1 )

then series number of B is Paschen( n = 3 ).

In this ${\left| \psi \right|^2}$ vs r graph, value of r is zero at only one point. So this orbital must have one radial node.

Here ${\left| \psi \right|^2} \ne 0$ at r = 0, so orbital should be s orbital. Then $l$ = 0.

We know,

No of radial node = n - $l$ - 1

$ \therefore $ n - $l$ - 1 = 1

$ \Rightarrow $ n - 0 - 1 = 1

$ \Rightarrow $ n = 2

Therefore orbital is 2s.

Atomic numbers of N, O, F and Na are 7, 8, 9 and 11 respectively.

Therefore, total number of electrons in each of N^3-, O^2–, F^– and Na⁺ is 10 and hence they are isoelectronic.

Probablity density = ${\psi ^2}$
From the graph you can see probablity density is maximum at the nucleus where r = 0.

$ \therefore $ Option A is correct.

(B) For hydrogen atom, electron can be found at a any distance distance from the nucleus.

$ \therefore $ Option B is correct.

(C) Distance a₀ from the nucleus = First orbital

$ \therefore $ n = 1 at first orbital

As Total energy, E_n = -13.6 $ \times $ ${{{Z^2}} \over {{n^2}}}$ eV

$ \therefore $ Total energy for hydrogen at n = 1 is,

E₁ = -13.6 $ \times $${{{1^2}} \over {{1^2}}}$ eV = -13.6 eV

So total energy is minimum at distance a₀ from the nucleus.

$ \therefore $ Option C is incorrect.

(D) Total energy, E_n = -13.6 $ \times $ ${{{Z^2}} \over {{n^2}}}$ eV

Kinetic energy, K.E = 13.6 $ \times $ ${{{Z^2}} \over {{n^2}}}$ eV

Potential energy, P.E = -27.2 $ \times $ ${{{Z^2}} \over {{n^2}}}$ eV

$ \therefore $ Magnitude of potential energy is double that of its kinetic energy.

$ \therefore $ Option D is correct.

We know,

$\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

$ \therefore $ For Lyman series,

${\Delta {{\overline v }_{Lyman}}}$ = ${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$

= $\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]$

= ${{1 \over 4}}$

$ \therefore $ For Balmer series,

${\Delta {{\overline v }_{Balmer}}}$ = ${\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}$

= $\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]$

= ${{1 \over 9}}$

$ \therefore $ ${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$ = ${{{1 \over 4}} \over {{1 \over 9}}}$ = ${{9 \over 4}}$

From photoelectric effect,

E = $\phi $ + KE

${{hc} \over \lambda } = \phi + {{{p^2}} \over {2m}}$ ................ (1)

Now when momentum = 1.5p then let wavelength = ${{\lambda _1}}$

$ \therefore $ ${{hc} \over {{\lambda _1}}} = \phi + {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Given,

kinetic energy(KE) of ejected photoelectron to be very high in comparison to work function($\phi $).

$ \therefore $ We can neglect work function($\phi $).

$ \therefore $ Equation (1) and (2) becomes,

${{hc} \over \lambda } = {{{p^2}} \over {2m}}$ ................ (1)

${{hc} \over {{\lambda _1}}} = {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Dividing (1) by (2) we get,

${{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}$

$ \Rightarrow $ ${{{\lambda _1}} \over \lambda } = {4 \over 9}$

$ \Rightarrow $ ${\lambda _1} = {4 \over 9}\lambda $

n = 4, l = 2,   4d orbital, n + l = 6

n = 3, l = 2,   3d orbital, n + l = 5

n = 4, l = 1,   4p orbital, n + l = 5

n = 3, l = 1,   3p orbital, n + l = 4

more is n + l value, more is energy

$ \therefore $ 3p < 3d < 4p < 4d

According to debroglie hypothesis,

2$\pi $r_n = n$\lambda $

$ \therefore $  $\lambda $ = ${{2\pi {r_n}} \over n}$

According to the question,

${{2\pi {r_n}} \over n}$ = 1.5 $\pi $a₀

We know,

r_n = 0.529 ${{{n^2}} \over Z}$

    = a₀ $ \times $ ${{{n^2}} \over Z}$

$ \therefore $  ${{2\pi } \over n} \times {{{a_0}{n^2}} \over Z}$ = 1.5 $\pi $ a₀

$ \Rightarrow $  ${n \over Z}$ = 0.75

Sun emits UV-radiations, which have the wavelength range from 1 nm to 400 nm.

E = $\phi $ + K.E

$ \Rightarrow $ h$\nu $ = $\phi $ + ${1 \over 2}m{v^2}$

$ \Rightarrow $ $\phi $ = $h\nu - {1 \over 2}m{v^2}$

= ${{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}}}} - {1 \over 2} \times 9 \times {10^{ - 31}} \times {\left( {6 \times {{10}^5}} \right)^2}$

= 3.35 $ \times $ 10^-19 J

$ \Rightarrow $ $\phi $ = ${{3.35 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$ eV

= 2.0934 eV $ \simeq $ 2.1 eV

By photoelectric effect

KE = h$\gamma $ - h$\gamma $_o ....(1)

de broglie wavelength,

$\lambda $ = ${h \over {mv}}$ = ${h \over {\sqrt {2m \times K.E} }}$ ...(2)

Using equation (1) and (2), we get

$\lambda $ = ${h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}$

$ \therefore $ $\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}$

Given, R_H = 1 $ \times $ 10⁵ cm^–1

$ \Rightarrow $ ${1 \over {{R_H}}}$ = 10^-5 cm

$ \Rightarrow $ ${1 \over {{R_H}}}$ = 10^-7 cm $ \times $ 100

$ \Rightarrow $ ${1 \over {{R_H}}}$ = 100 nm

We know,

${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$

$ \Rightarrow $ $\lambda $ = ${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

[For H atom Z = 1]

$ \Rightarrow $ $\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

$ \Rightarrow $ $\lambda $ = ${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$

Given, $\lambda $ = 900 nm

$ \therefore $ ${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$ = 900

$ \Rightarrow $ ${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$

By checking each options you can see

when n_L = 3 and n_H = $\infty $ then

${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$

$ \therefore $ Option C is correct.

(E)_n^th = (E_GND)_H . ${{{Z^2}} \over {{n^2}}}$

E_3^rd (He⁺) = ($-$13.6eV) . ${{{2^2}} \over {{3^2}}}$ = $-$ 6.04 eV

E = W + ${1 \over 2}$mv²

K.E. = hv $-$ 4v₀

K.E. = hv + ($-$ hv₀)

y = mx + $\underline C $

(a) Angular momentum, $mvr = {{nh} \over {2\pi }}$

$ \Rightarrow mvr \propto n$ $\propto$ distance from the nucleus

(b) This statement is incorrect as size of an orbit $\propto$ Azimuthal quantum number (l) ($\because$ n = constant)

(c) This statement is incorrect as at ground state,

n = 1, l = 0

$\Rightarrow$ Orbital angular momentum (wave mechanics)

$ = \sqrt {l(l + 1)} {h \over {2\pi }} = 0$ [$\because$ $l = 0$]

(d) The plot of $\psi $ vs $ r $ for various azimuthal quantum numbers, shows peak shifting towards higher $ r $ value.
This statement is typically true. The radial distribution functions of atomic orbitals (which can be related to $ |\psi|^2 $ vs $ r $) for higher azimuthal quantum numbers generally have their peaks at larger values of the radial distance $ r $ from the nucleus. This is because with higher $ l $ values, the electrons are more likely to be found at greater distances due to the higher angular momentum.

The given plot is

In transition metal contains d orbital, and in d orbital maximum no of unpaired electron possible = 5.

Spin only magnetic moment,

${\mu _{spin}} = \sqrt {n\left( {n + 2} \right)} $    B. M

here n $=$ Number of unpaired electrons.

$ \therefore $    ${\mu _{spin}} = \sqrt {5\left( {5 + 2} \right)} $    B. M

$=$ 5.92   B. M

We know,

$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$

here n_i $=$ 8 and n_f $=$ n

Z $=$ 1 for hydrogen

$ \therefore $   $\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$

So, graph is $ \to $



So, graph will be linear with slope R_H.

(III) Kinetic energy of the electron in nth orbit,

$K.E. = + 13.6 \times {{{Z^2}} \over {{n^2}}}$ or $K.E. \propto {1 \over {{n^2}}}$ or $K.E. \propto {n^{ - 2}}$

From List-II, correct match is (III, P)

(IV) Potential energy of the electron in the nth orbit,

$P.E. = - 2 \times 13.6 \times {{{Z^2}} \over {{n^2}}} \Rightarrow P.E. \propto {1 \over {{n^2}}} \Rightarrow P.E. \propto {n^{ - 2}}$

From List-II, correct match is (IV, P).

Hence, correct matching from List-I and List-II on the basis of given options is (III, P).

(I) Radius of the nth orbit, r = $0.529 \times {{{n^2}} \over Z}$

Here, $r \propto {n^2}$

From List-II, correct match is (I, T)

(II) Angular momentum of the electron, $mvr = {{nh} \over {2\pi }}$ or $mvr \propto n$

From List - II, correct match (II, S)

Hence, correct matching from List-I and List-II on the basis of given options is (I, T).

Given, ground state energy of hydrogen atom = $-$13.6 eV

Energy of He⁺ = $-$3.4 eV, Z = 2

Energy of He⁺, E = $ - {{13.6 \times {Z^2}} \over {{n^2}}}eV$

$ - 3.4eV = {{ - 13.6 \times {{(2)}^2}} \over {{n^2}}}$

$ \Rightarrow $ n = $\sqrt {{{ - 13.6 \times 4} \over {34}}} $ $ \Rightarrow $ n = 4

Given, azimuthal quantum number (l) = 2 (d $-$ subshell)

Magnetic quantum number (m) = 0

$ \therefore $ Angular nodes (l) = 2

Radial node = n $-$ l $-$ 1 = 4 $-$ 2 $-$ 1 = 1

nl = 4d state

Hence, options (a), (c) are correct.

When a black body is heated, black body emit high energy radiation, from higher wavelength to lower wavelength.

Radius    r = 0.529 $ \times $ ${{{n^2}} \over z}$

In first Bohr orbit of hydrogen atom radius,

r = 0.529 $ \times $ ${{{1^2}} \over 1}$ = 0.529 $\mathop A\limits^ \circ $

Angular momentum,

mvr = ${{nh} \over {2\pi }}$

$\therefore\,\,\,$ for    n = 1,

    mvr = ${h \over {2\pi }}$

$ \Rightarrow $$\,\,\,$ 2$\pi $r = ${h \over {mv}}$ = $\lambda $   [ as   $\lambda $ = ${h \over {mv}}$]

$\therefore\,\,\,$ $\lambda $ = 2$\pi $r = 2 $\pi $ $ \times $ 0.5 29

$\lambda $ = 250 nm = 2500 $\mathop A\limits^ \circ $

E = ${{hc} \over \lambda }$ = ${{12400} \over {2500}}$ = 4.96 eV

K. E = 0.5 eV

As, $\,\,\,\,\,\,$ K. E = E $-$ $\omega $₀

$ \Rightarrow $ $\,\,\,$ 0.5 = 4.96 $-$ ${\omega _0}$

$ \Rightarrow $ $\,\,\,$ ${\omega _0}$ = 4.46 eV $ \simeq $ 4.5 eV

Here electron is coming to the orbital of radius 211.6 pm. Now we have to find which series have radius of orbital 211.6 pm.

We know,

Radius, r = $0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ $

Given, r = 211.6 pm = 211.6 $ \times $ 10^-12 m

and Z = 1 for hydrogen atom

$ \therefore $ 211.6 $ \times $ 10^-12 = $0.529 \times {{{n^2}} \over 1}$ $ \times $ 10^-10

$ \Rightarrow $ n = 2

As n = 2 so the series is Balmer Series.

Note :

(1)   In Lyman Series, transition happens in n = 1 state
from n = 2, 3, . . . . . $ \propto $

(2)   In Balmer Series, transition happens in n = 2 state
from n = 3, 4, . . . . . $ \propto $

(3)   In Paschen Series, transition happens in n = 3 state
from n = 4, 5, . . . . . $ \propto $

(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $ \propto $

(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $ \propto $

We know,

     ${1 \over \lambda }$ = Rz² (${1 \over {n_1^2}}$ $-$ ${1 \over {n_2^2}}$)

The shortest wavelength of hydrogen atom in Lyman series is from n₁ = 1 to n₂ = $ \propto $

$\therefore\,\,\,$ ${1 \over A}$ = 1² R (${1 \over {{1^2}}}$ $-$ ${1 \over {{ \propto ^2}}}$)    [for hydrogen, z = 1]

$ \Rightarrow $ $\,\,\,$ ${1 \over A}$ = R

The longest wavelength in pascal series of He⁺ is from n₁ = 3 to n₂ = 4

For He⁺, z = 2.

${1 \over \lambda }$ = RZ² (${1 \over {n_1^2}}$ $-$ ${1 \over {n_2^2}}$)

$ \Rightarrow $$\,\,\,$ ${1 \over \lambda }$ = ${1 \over A}$ (2)² (${1 \over {{3^2}}}$ $-$ ${1 \over {{4^2}}}$)

$ \Rightarrow $$\,\,\,$ ${1 \over \lambda }$ = ${4 \over A}$ $\left( {{7 \over {16 \times 9}}} \right)$ = ${7 \over {36A}}$

$ \Rightarrow $$\,\,\,$ $\lambda $ = ${{36A} \over 7}$

Radius of an atom in n^th orbit,

r_n = 0.529 $ \times {{{n^2}} \over Z}$

Here n = 2

and for hydrogen atom, atomic number (Z) = 1

$\therefore$ r₂ = 0.529$ \times {{{2^2}} \over 1}$ = 2.12 $\mathop {\rm A}\limits^o $

O^2- , F^- , Na⁺ and Mg²⁺, all have 10 electrons each.

- Option (A) : Incorrect. The total number of radial nodes is given by $n-l-1$. For $1 s$ orbital, the radial node is zero (i.e. $1-0-1=0$ ).

- Option (B) : Incorrect. $d_{z^2}$ orbital has two lobes that point in opposite directions along $z$-axis plus a doughnut-shaped ring of electron density around the centre that lies in the $x y$-plane. Therefore, it does not have any nodal plane.

- Option (C) : Correct. The total number of radial nodes is given by $n-l-1$.

For $2 s$ : Radial nodes $=2-0-1=1$

The orbital 2s of hydrogen like species can be diagrammatically represented as :


There is a spherical region of zero electron density between $1 s$ and the $2 s$ orbital of hydrogen like species. This spherical region is called the node. Only one radial node exist between $1 s$ and $2 s$.

The plot of wave function for $2 s$ verses radial distance from the nucleus is represented as



Point $P$ represents the region of zero electron density at a distance $r$ from the nucleus. This represents spherical node.

- Option (D) : Incorrect. The probability density at nucleus is for $2p$ orbital is zero.

The ion $\mathrm{He}^{+}$having 1 electron (in $1 s$ orbital) is hydrogen like species; hence, its $1 s$ orbital will have:

(i) Wave function same as that of hydrogen, i.e.,

$ \psi_{1,0,0}=\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

(ii) The probability density for the electron in the $1 s$ orbital is represented as:


For $1 s$ orbital probability density is maximum at the nucleus.

(iii) Probability density for $2 s$ orbitals is given by

$ =\frac{1}{2 a_0^{3 / 2}}\left(1-\frac{r}{2 a_0}\right) e^{-r / 2 a_0} $

Probability density $\propto \frac{1}{a_0^3}$

(iv) For electron in the $1 s$ orbital, the wave function of electron is not the function of angular wave function i.e., $\theta$ and $\phi$. Hence, 1 s orbital cannot have the wave function

$ \psi_{n, l, m_l} \propto\left(\frac{Z}{a_0}\right)^{5 / 2} r e^{\left(-Z r / 2 a_0\right) \cos \theta} $

(i) Wave function for 1s orbital of hydrogen atom with $n, l$ and $m_l$ as quantum numbers is given

$ \psi_{n, l, m_l}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} $

Where, $\mathrm{Z}$ is atomic number of the element, $r$ is the variable distance of electron from the nucleus, $a_0$ is the Bohr's radius

For $1 \mathrm{~s}$ orbital, $n=1, l=0, m_l=0$ and wave function is represented as :

$ \begin{aligned} & \psi_{1,0,0}=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \\\\ \Rightarrow \psi_{1,0,0} & \propto\left(\frac{Z}{a_0}\right)^{3 / 2} e^{-Z r / a_0} \end{aligned} $

(ii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as :

$ \begin{aligned} & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{1}{4^2}-\frac{1}{2^2}\right) \\\\ & \mathrm{E}_{2 \rightarrow 4} \propto-\mathrm{R}\left(\frac{-3}{16}\right)=\frac{3 R}{16} \end{aligned} $

(iii) Energy of excitation of electron from $n_1=2$ to $n_2=4$ is given as:

$ \begin{aligned} \mathrm{E}_{2 \rightarrow 6} & \propto-\mathrm{R}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\\\ \mathrm{E}_{2 \rightarrow 6} & \propto-R\left(\frac{1}{6^2}-\frac{1}{2^2}\right) \\\\ & =-\left(\frac{1}{36}-\frac{1}{4}\right) R \\\\ E_{2 \rightarrow 6} & \propto \frac{+8 R}{36} \end{aligned} $

The ratio of the energy of transition in the two cases is given by :

$ \begin{aligned} & \frac{E_{2 \rightarrow 4}}{E_{2 \rightarrow 6}}=\frac{3 R}{16} \times \frac{6}{8 R}=\frac{27}{32} \\\\ & E_{2 \rightarrow 4}=\frac{27}{32} \times E_{2 \rightarrow 6} \end{aligned} $

the energy of transition from 2 to 4 excited state is $\frac{27}{32}$ times the energy of transition from 2 to 6 excited state.

Total number of orbitals associated with principal quantum number n is = n².

Here, n = 5.

$ \therefore $ The total number of orbitals associated with the principal quantum number 5 is = 5² = 25

We know, Kinetic Energy (K_E) = ${1 \over 2}m{v^2}$

$\therefore$ mv² = 2K_E

$ \Rightarrow $ m²v² = 2mK_E

$ \Rightarrow $ mv = $\sqrt {2m{K_E}} $

For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be

K_E = eV

$\therefore$ m = $\sqrt {2meV} $

We know de-Broglie wavelength for an electron$\left( \lambda \right)$ = ${h \over p} = {h \over {mv}}$

$\therefore$ ${h \over \lambda }$ = mv = $\sqrt {2meV} $

The probability of finding an electron of hydrogen atom in a spherical shell of infinitesimal thickness, $d r$, at a distance $r$ from the nucleus, with volume $d V=4 \pi r^2 d r$ is

$ P=\psi^2 \cdot d V=\psi^2 \cdot 4 \pi r^2 d r=R^2(r) 4 \pi r^2 d r $

Here, $R^2(r)$ is the radial density function.

For $1 s$ subshell, $n=1, l=0$ and $n-l-1=0$. Thus, the number of radial and angular nodes $=0$. For 1s orbital, the probability of finding an electron will be maximum at near to the the nucleus, as 1s orbital is the nearest to the nucleus as it is the lowest orbital in terms of energy. Therefore, the plot of radial probability $P=R^2(r) 4 \pi r^2$ versus $r$ is as follows :

Energy(E_n) = $ - {{13.6{Z^2}} \over {{n^2}}}eV$

where n = 1, 2, 3, 4, ......

For hydrogen Z = 1 and excited state starts from n = 2

So putting n = 2

E₂ = $ - {{13.6} \over {{2^2}}}eV$ = -3.4 eV

The electronic configuration of Rubidium (Rb = 37):
1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹

As you can see last electron or valence electron enterin 5s subshell

So, the quantum numbers are n = 5, $l$ = 0, m = 0, s = $ \pm {1 \over 2}$

Given $E$ = $- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$
                = $ - 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n₂ = 2 and n₁ = 1.
And for H, Z = 1
$\therefore$ $E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$
           = 1.6335 $ \times $ 10^-18 J

We know $E$ = ${{hc} \over \lambda }$
$\therefore$ ${{hc} \over \lambda }$ = 1.6335 $ \times $ 10^-18
then $\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$
             $ = 1.214 \times {10^{ - 7}}m$