Structure of Atom Test

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Structure of Atom Test - 70

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Question 1
1 / -0

The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given, $$r=0.53\overset{o}{A}$$).

A
$$6\times 10^5$$Hz

B
$$2.5\times 10^{12}$$Hz

C
$$1.9\times 10^{13}$$Hz

D
$$6.5\times 10^{15}$$Hz

Solution

From Bohr's theory
$$mvr=\displaystyle\frac{nh}{2\pi}$$
For first orbit $$n=1$$
$$m(r\omega )r=\displaystyle\frac{1\times h}{2\pi}$$
$$\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}$$
$$\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}$$
$$=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}$$.
=$$6.5\times10^{15} Hz$$
Question 2
1 / -0

The highest energy in Balmer series, in the emission spectra of hydrogen is represented by:
$$\left({R}_{H}=109737 {cm}^{-1}\right)$$

A
$$4389.48 {cm}^{-1}$$

B
$$2194.74 {cm}^{-1}$$

C
$$5486.85 {cm}^{-1}$$

D
$$27434.25 {cm}^{-1}$$

Solution

$$\bar{v}=\dfrac{1}{\lambda}=R_H{Z}^{2}\left[\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right]$$

$$=109737 \times {1}^{2} \left[\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right] = 27434.25 {cm}^{-1}$$

(series limit will be of highest energy)
Question 3
1 / -0

The energy of second Bohr orbit of the hydrogen atom is $$-328kJ\ { mol }^{ -1 }$$. Hence, the energy of fourth Bohr orbit would be:

A
$$-41kJ\ { mol }^{ -1 }$$

B
$$-1312kJ\ { mol }^{ -1 }$$

C
$$-164kJ\ { mol }^{ -1 }$$

D
$$-82kJ\ { mol }^{ -1 }$$

Solution

$$ \displaystyle E \propto \dfrac { 1 }{ n^2 }$$

$$ \displaystyle \dfrac { E_4 }{ E_2 } = \left(\dfrac { n_2 }{ n_4 } \right)^2$$

$$ \displaystyle { E_4 }={-328 \: kJ/mol } \times \left(\dfrac { n_2 }{ n_4 } \right)^2$$

$$ \displaystyle { E_4 }={-328 \: kJ/mol } \times \left(\dfrac { 2 }{ 4 } \right)^2$$

$$ \displaystyle { E_4 }={-82 \: kJ/mol }$$

Hence, the energy of fourth Bohr orbit would be $$-82kJ\ { mol }^{ -1 }$$.
Question 4
1 / -0

Which of the following orbitals has the highest energy?

A
$$5d$$

B
$$5f$$

C
$$6s$$

D
$$6p$$

Solution

The orbital energies are compared with the help of $$n+l$$ rule. The higher value of $$n+l$$, higher is the energy of the orbital.

$$5d = n+l =7$$
$$5f = n+l =8$$
$$6s = n+l =6$$
$$6p = n+l =7$$

The $$n+l$$ value is greatest for 5f orbital.
Hence it has the highest energy among the given options.
Question 5
1 / -0

The spectral lines corresponding to the radiation emitted by an electron jumping from higher orbits to first orbit belong to:

A
Paschen series

B
Balmer series

C
Lyman series

D
None of these

Solution

The spectral lines corresponding to the radiation emitted by an electron jumping from higher orbits to first orbit belong to Lyman series.
Lyman series corresponds to transitions to $$\displaystyle n = 1$$ level.
Balmer series corresponds to transitions to $$\displaystyle n = 2$$ level.
Paschen series corresponds to transitions to $$\displaystyle n = 3$$ level.
Pfund series corresponds to transitions to $$\displaystyle n = 4$$ level.
Question 6
1 / -0

If the shortest wavelength of the spectral line of $${ He }^{ + }$$ in Lyman series is $$x$$, then longest wavelength of the line in Balmer series of $${ Li }^{ 2+ }$$ is:

A
$$\cfrac { 5x }{ 4 } $$

B
$$\cfrac { 4x }{ 5 } $$

C
$$\cfrac { 16x }{ 5 } $$

D
$$9x$$

Solution

Shortest wavelength $$={ n }_{ \infty }\rightarrow { n }_{ 1 }$$

$$\cfrac { 1 }{ { \lambda }_{ min } } ={ R }_{ H }\cfrac { { Z }^{ 2 } }{ { n }_{ 1 }^{ 2 } } $$

for $${He}^{+},Z=2$$

$$\cfrac { 1 }{ { \lambda }_{ min } } ={ R }_{ H }\cfrac { { 2 }^{ 2 } }{ { 1 }_{ }^{ 2 } } \quad $$ or $${ R }_{ H }=\cfrac { 1 }{ 4x } $$

for the longest wavelength in Balmer series of $${ Li }^{ 2+ },Z=3,{ n }_{ 1 }=2,{ n }_{ 2 }=3$$

$$\cfrac { 1 }{ { \lambda }_{ max } } ={ R }_{ H }{ Z }^{ 2 }\left( \cfrac { 1 }{ { n }_{ 1 }^{ 2 } } -\cfrac { 1 }{ { n }_{ 2 }^{ 2 } } \right) $$

$$\cfrac { 1 }{ { \lambda }_{ max } } =\cfrac { 1 }{ 4x } { (3) }^{ 2 }\left( \cfrac { 1 }{ 4 } -\cfrac { 1 }{ 9 } \right) $$

$$\cfrac { 1 }{ { \lambda }_{ max } } =\cfrac { 9 }{ 4x } \times \cfrac { 5 }{ 36 } $$

$$\quad { \lambda }_{ max }=\cfrac { 16x }{ 5 } $$
Question 7
1 / -0

The fast emission line on hydrogen atomic spectrum in the Balmer series appears at:

[$$R =$$ Rydberg constant]

A
$$\dfrac{5R}{36}cm^{-1}$$

B
$$\dfrac{3R}{4}cm^{-1}$$

C
$$\dfrac{7R}{144}cm^{-1}$$

D
$$\dfrac{9R}{400}cm^{-1}$$

Solution

Rydberg - Riz equation is :

$$\bar{v}=R_H\left [ \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2} \right ] \,\,\, (Where \, n_1= 2, n_2=3)$$

$$\therefore \bar{v}=R_H\left [ \dfrac{9-4}{64} \right ] =\dfrac{5R} {36} cm^{-1}$$
Question 8
1 / -0

The magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is:
[$$a_0$$ is Bohr radius].

A
$$\dfrac{h^2}{4 \pi^2 ma_o^2}$$

B
$$\dfrac{h^2}{16 \pi^2 ma_o^2}$$

C
$$\dfrac{h^2}{32 \pi^2 ma_o^2}$$

D
$$\dfrac{h^2}{64 \pi^2 ma_o^2}$$

Solution

Solution:- (B) $$\cfrac{{h}^{2}}{16 {\pi}^{2} m a_o^2}$$
According to Bohr's model of an atom, the potential energy of electron having charge $$-e$$ is given as-
$$P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{e}{r} \left( -e \right)$$
$$\Rightarrow P.E. =- \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{r}$$
$$\because r = \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}}$$
$$\therefore P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{\left( \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}} \right)}$$
$$\Rightarrow P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m {e}^{4}}{{n}^{2} {h}^{2}}$$
Now,
$${a}_{0} = \cfrac{{\epsilon}_{0} {h}^{2}}{\pi m {e}^{2}}$$
$$\Rightarrow {e}^{4} = \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}}$$
Therefore,
$$\therefore P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m}{{n}^{2} {h}^{2}} \left( \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}} \right)$$
$$\Rightarrow P.E. = \cfrac{{h}^{2}}{4 {\pi}^{2} {n}^{2} m {{a}_{0}}^{2}}$$
For second orbit of hydrogen atom,
$$n = 2$$
$$\Rightarrow P.E. = \cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}$$
Hence the magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is $$\cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}$$.
Question 9
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The minimum wavelength absorbed by electron initially present in $$(n_1+1)$$ shell of $$Li^{+2}$$ ion:(during its excitation)

A
$$\dfrac{n^2_1}{9R}$$

B
$$\dfrac{(n_1+1)^2}{9R}$$

C
$$\dfrac{n_1(n^1_1+1)^2}{2R}$$

D
$$\dfrac{n^2_1}{R \times 4^2}$$

Solution

Minimum wave length means maximum energy
So, $$n_2= \infty$$
$$\begin{array}{l} \frac { 1 }{ \lambda } ={ R_{ H } }{ Z^{ 2 } }\left( { \frac { 1 }{ { { { \left( { { n_{ 1 } }+1 } \right) }^{ 2 } } } } -\frac { 1 }{ { { { \left( \infty \right) }^{ 2 } } } } } \right) \Rightarrow { R_{ H } }\times { \left( 3 \right) ^{ 2 } }\left( { \frac { 1 }{ { { { \left( { { n_{ 1 } }+1 } \right) }^{ 2 } } } } } \right) \\ \lambda =\frac { { { { \left( { { n_{ 1 } }+1 } \right) }^{ 2 } } } }{ { 9{ R } } } \end{array}$$
Option B is correct.
Question 10
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Calculate the wavelength of the first line of $$He^{+}$$ ion spectral series whose interval with extreme lines is:
$$\dfrac{1}{\lambda_{1}} - \dfrac{1}{\lambda_{2}} = 2.7541 \times 10^{4} cm^{-1}$$

A
$$4869 \overset{o}{A}$$

B
$$1651 \overset{o}{A}$$

C
$$3039 \overset{o}{A}$$

D
$$2175 \overset{o}{A}$$

Solution

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Structure of Atom Test - 70

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Structure of Atom Test - 70

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Question 1

$$6\times 10^5$$Hz

$$2.5\times 10^{12}$$Hz

$$1.9\times 10^{13}$$Hz

$$6.5\times 10^{15}$$Hz

Solution

Question 2

$$4389.48 {cm}^{-1}$$

$$2194.74 {cm}^{-1}$$

$$5486.85 {cm}^{-1}$$

$$27434.25 {cm}^{-1}$$

Solution

Question 3

$$-41kJ\ { mol }^{ -1 }$$

$$-1312kJ\ { mol }^{ -1 }$$

$$-164kJ\ { mol }^{ -1 }$$

$$-82kJ\ { mol }^{ -1 }$$

Solution

Question 4

$$5d$$

$$5f$$

$$6s$$

$$6p$$

Solution

Question 5

Paschen series

Balmer series

Lyman series

None of these

Solution

Question 6

$$\cfrac { 5x }{ 4 } $$

$$\cfrac { 4x }{ 5 } $$

$$\cfrac { 16x }{ 5 } $$

$$9x$$

Solution

Question 7

$$\dfrac{5R}{36}cm^{-1}$$

$$\dfrac{3R}{4}cm^{-1}$$

$$\dfrac{7R}{144}cm^{-1}$$

$$\dfrac{9R}{400}cm^{-1}$$

Solution

Question 8

$$\dfrac{h^2}{4 \pi^2 ma_o^2}$$

$$\dfrac{h^2}{16 \pi^2 ma_o^2}$$

$$\dfrac{h^2}{32 \pi^2 ma_o^2}$$

$$\dfrac{h^2}{64 \pi^2 ma_o^2}$$

Solution

Question 9

$$\dfrac{n^2_1}{9R}$$

$$\dfrac{(n_1+1)^2}{9R}$$

$$\dfrac{n_1(n^1_1+1)^2}{2R}$$

$$\dfrac{n^2_1}{R \times 4^2}$$

Solution

Question 10

$$4869 \overset{o}{A}$$

$$1651 \overset{o}{A}$$

$$3039 \overset{o}{A}$$

$$2175 \overset{o}{A}$$

Solution

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