Atoms Test

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Atoms Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

The radius of the smaller electron orbit in hydrogen-like ion is $$\dfrac{0.51\times 10^{-10}}{4}$$m, then it is?

A
hydrogen atom

B
$$He^+$$

C
$$Li^+$$

D
$$Be^{3+}$$

Solution

For hydrogen like atom, the radius of nth orbit
$$r^z_n=\displaystyle\frac{n^2}{Z}a_0$$
Here, $$a_0=0.51\times 10^{-10}$$m
$$\therefore r^z_n=\displaystyle\frac{0.51\times 10^{-10}}{4}$$m
In the ground state, $$n=1$$
$$\therefore \displaystyle\frac{0.51\times 10^{-10}}{4}=\frac{1^2}{z}\times 0.51\times 10^{-10}$$
$$\therefore Z=4$$
So, the atom is triply ionised beryllium$$(Be^{3+})$$.
Question 2
1 / -0

A single electron orbits around a stationary nucleus of charge $$+Ze$$, where $$Z$$ is a constant and $$e$$ is the magnitude of the electronic charge. It required $$47.2\ eV$$ to excite the electron from the second Bohr orbit to the third Bohr orbit.
Find the energy required to excite the electron from the third to the fourth Bohr orbit.

A
$$18\ eV$$

B
$$16.53\ eV$$

C
$$19\ eV$$

D
$$21\ eV$$

Solution

$$ \begin{array}{l} \Delta E=13.6\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) z^{2} \\ \text { given } n_{1}=2, n_{2}=3 \\ 47 \cdot 2=13 \cdot 6 z^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\ z^{2}=25 \\ z=5 \end{array} $$
$$ \begin{array}{l} \text { now, putting } n_{1}=3, n_{2}=4 \\ \Delta E=13.6 \times 2^{2} \times\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right) \mathrm{} \\ =13.6 \times 25 \times \frac{7}{16 \times 9} \\ =16.53 \mathrm{eV} \end{array} $$
$$\text{Correct B}$$
Question 3
1 / -0

In an experiment on photoelectric effect photons of wavelength $$300\ nm$$ eject electrons from a metal of work function $$2.25\ eV$$. A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom.

A
$$n = 2$$ to $$n = 1$$ state

B
$$n = 3$$ to $$n = 1$$ state

C
$$n = 3$$ to $$n = 2$$ state

D
$$n = 4$$ to $$n = 3$$ state

Solution

Given $$\phi =2.25eV\\ \lambda =300\times { 100 }^{ -3 }m$$
Energy of the given photon,
$$E=h\upsilon \\ =\frac { hc }{ \lambda } =\frac { 663\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 300\times { 10 }^{ -9 } } =6.63\times { 10 }^{ -19 }J=4.13eV$$
Now,
$$E=\frac { hc }{ \lambda } -\phi =4.137-2.25=1.88eV\rightarrow (1)$$
So, 1.88eV energy is used to jump from one orbit to another orbit by electron.
Therefore, energy of different orbital of hydrogen
$$\Rightarrow n=1\qquad \qquad \qquad \quad 2\qquad \quad \quad 3\qquad \quad \quad 4\\ \quad \quad E=-13.6ev\qquad -3.4\qquad -1.51\quad \quad -0.85$$
And from statement (1),
$$1.18={ E }_{ i }-{ E }_{ f }\quad \\ $$
{$${ E }_{ i }\& { E }_{ f }\quad $$ are energy of initial and final orbit}
And also from the above table, we can observe that,
$${ E }_{ i }-{ E }_{ f }=(-1.51)-(-3.4)=1.89eV\\ ={ { E }_{ 2 }-{ E }_{ 3 } }\quad \\ $$
That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 300nm wavelength hydrogen electron can jump from 3 to 2.
Question 4
1 / -0

Directions For Questions

The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

...view full instructions

A diatomic molecules has moment of inertia $$I$$. By Bohr's quantization condition its rotational energy in the $$n^{th}$$ level $$(n = 0$$ is not allowed) is

A
$$\dfrac {1}{n^{2}}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

B
$$\dfrac {1}{n}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

C
$$n\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

D
$$n^{2}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

Solution
Question 5
1 / -0

A mixture of $$2$$ moles of helium gas (atomic mass $$=4$$ amu) and $$1$$ mole of argon gas (atomic mass $$=40$$ amu) is kept at $$300$$ K in a container. The ratio of the rms speeds $$\left(\dfrac{v_{rms}(helium)}{v_{rms}(argon)}\right)$$ is?

A
$$0.32$$

B
$$0.45$$

C
$$2.24$$

D
$$3.16$$

Solution

$$V_{rms}=\sqrt{\cfrac{3RT}{M}}$$

$$V_{rms}$$ for helium $$=\sqrt{\cfrac{3RT}{4}}$$

$$V_{rms}$$ for argon $$=\sqrt{\cfrac{3RT}{40}}$$

$$\cfrac{V_{rms}(helium)}{V_{rms}(Argon)}=\sqrt{10}$$
$$=3.16$$
Question 6
1 / -0

Which of the following statement is wrong?

A
$$\Psi_s$$ = - $$\pi$$

B
DPD = -OP + TP

C
$$\Psi_w$$ = $$\Psi_s$$ + $$\Psi_p$$

D
$$\Psi_w$$ = -DPD
Question 7
1 / -0

A single electron orbits around a stationary nucleus of charge $$+Ze$$, where $$Z$$ is a constant and $$e$$ is the magnitude of the electronic charge. It required $$47.2\ eV$$ to excite the electron from the second Bohr orbit to the third Bohr orbit. Find the radius of the first Bohr orbit.
(The ionization energy of hydrogen atom $$13.6\ eV$$, Bohr radius $$= 5.3\times 10^{-11} m$$, speed of light $$= 3\times 10^{8} m/s$$, Planck's constant $$= 6.6\times 10^{-34} J-sec)$$.

A
$$1\times 10^{-11}m$$

B
$$3\times 10^{-11}m$$

C
$$5\times 10^{-11}m$$

D
$$8\times 10^{-11}m$$

Solution
Question 8
1 / -0

In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed. The wave numbers which may be observed in Balmer series is then?

A
$$\dfrac{8R}{9}, \dfrac{5R}{36}$$

B
$$\dfrac{3R}{16}, \dfrac{8R}{9}$$

C
$$\dfrac{5R}{36}, \dfrac{3R}{16}$$

D
$$\dfrac{3R}{4}, \dfrac{3R}{16}$$

Solution

$$Given:$$ In certain electronic transition from quantum level n to ground state in atomic hydrogen in one or more steps no line belonging to Brackett series is observed.
$$Solution:$$
The $$n$$ should be equal to 4 ; otherwise if $$n>4$$,
then Bracket series will also be noticed.
Total no.of lines formed $$=\mathbf{n}(\mathbf{n}-1) / 2=4 \times 3 / 2=6$$
No. of lines formed will be due to transition
between these energy levels $$\mathbf{n}=4$$ to $$\mathbf{n}=\mathbf{3}$$
$$\mathbf{n}=4 \operatorname{ton}=2$$
$$\mathbf{n}=4$$ to $$\mathbf{n}=\mathbf{1}$$
$$\mathbf{n}=3 \operatorname{ton}=2$$
$$\mathbf{n}=3 \operatorname{ton}=1$$
$$\mathbf{n}=2$$ to $$\mathbf{n}=1$$
Balmer series line obtained when any electron comes to second orbit from outer orbit so
out of these six lines, only two lines belongs to Balmer series.
i.e., from $$n=4$$ to $$n=2$$ and
$$\mathbf{n}=3$$ to $$\mathbf{n}=2$$
For these lines wave number is,
$$(\mathrm{i}) \overline{\mathbf{v}{1}}=\frac{\mathbf{1}}{\lambda}=\mathbf{R}{\mathbf{H}}\left[\frac{\mathbf{1}}{\mathbf{2}^{2}}-\frac{\mathbf{1}}{\mathbf{4}^{2}}\right]=\frac{\mathbf{1 2}}{\mathbf{6 4}} \mathbf{R}{\mathbf{H}}=\frac{\mathbf{3}}{\mathbf{1 6}} \mathbf{R}{\mathbf{H}}$$
$$(\mathrm{ii}) \overline{\mathrm{v}{2}}=\frac{1}{\lambda} \mathbf{R}{\mathrm{H}}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5}{36} \mathbf{R}_{\mathrm{H}}$$
$$So,the$$ $$correct$$ $$option:C$$
Question 9
1 / -0

If velocity of $$\alpha - $$particle is made $$1/3$$ times, then % variation in distance of closest approach will be: (with respect to initial)

A
300%

B
600%

C
800%

D
80%

Solution

The distance of closest approach of a particle is given by,
$$r=\dfrac { 4KZ{ e }^{ 2 } }{ m{ v }^{ 2 } } $$
Now, in second case, when $$v\rightarrow \dfrac { v }{ 3 } $$, distance of closed approach will be,
$${ r }^{ ' }=\dfrac { 4kZ{ e }^{ 2 } }{ m{ \left( \dfrac { v }{ 3 } \right) }^{ 2 } } =\dfrac { 4kZ{ e }^{ 2 } }{ m{ { v } }^{ 2 } } =9r$$
If r represents $$100$$%
then $$9r=900$$%.
Percentage variation$$=900-100=800$$%
Question 10
1 / -0

In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as
$$\quad F=\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 } } \left( \cfrac { 1 }{ { r }^{ 2 } } +\cfrac { \beta }{ { r }^{ 3 } } \right) $$, where $$\beta$$ is a constant. For this atom, the radius of the $$n$$th orbit in terms of the Bohr radius $$\left( { a }_{ 0 }=\cfrac { { \varepsilon }_{ 0 }{ h }^{ 2 } }{ m\pi { e }^{ 2 } } \right) $$ is:

A
$${ r }_{ n }={ a }_{ 0 }n-\beta $$

B
$${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }+\beta \quad $$

C
$${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }-\beta $$

D
$${ r }_{ n }={ a }_{ 0 }n+\beta $$

Solution

$$\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 } } \left( \cfrac { 1 }{ { r }^{ 3 } } -\cfrac { \beta }{ { r }^{ 3 } } \right) $$
$$\quad \therefore { v }^{ 2 }=\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{ } } +\cfrac { \beta }{ { r }^{ 2 } } \right) $$
From Bohr's hypothesis $$\quad v=\cfrac { nh }{ 2\pi mr } \quad \therefore \cfrac { { n }^{ 2 }{ hn }^{ 2 } }{ 4{ \pi }^{ 2 }{ m }^{ 2 }{ r }^{ 2 } } =\cfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }mr } \left( \cfrac { 1 }{ { r }^{ } } +\cfrac { \beta }{ { r }^{ 2 } } \right) $$
$$\therefore r+\beta ={ a }_{ 0 }{ n }^{ 2 }\quad $$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 70

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Atoms Test - 70

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SHARING IS CARING

Question 1

hydrogen atom

$$He^+$$

$$Li^+$$

$$Be^{3+}$$

Solution

Question 2

$$18\ eV$$

$$16.53\ eV$$

$$19\ eV$$

$$21\ eV$$

Solution

Question 3

$$n = 2$$ to $$n = 1$$ state

$$n = 3$$ to $$n = 1$$ state

$$n = 3$$ to $$n = 2$$ state

$$n = 4$$ to $$n = 3$$ state

Solution

Question 4

$$\dfrac {1}{n^{2}}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

$$\dfrac {1}{n}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

$$n\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

$$n^{2}\left (\dfrac {h^{2}}{8\pi^{2}I}\right )$$

Solution

Question 5

$$0.32$$

$$0.45$$

$$2.24$$

$$3.16$$

Solution

Question 6

$$\Psi_s$$ = - $$\pi$$

DPD = -OP + TP

$$\Psi_w$$ = $$\Psi_s$$ + $$\Psi_p$$

$$\Psi_w$$ = -DPD

Question 7

$$1\times 10^{-11}m$$

$$3\times 10^{-11}m$$

$$5\times 10^{-11}m$$

$$8\times 10^{-11}m$$

Solution

Question 8

$$\dfrac{8R}{9}, \dfrac{5R}{36}$$

$$\dfrac{3R}{16}, \dfrac{8R}{9}$$

$$\dfrac{5R}{36}, \dfrac{3R}{16}$$

$$\dfrac{3R}{4}, \dfrac{3R}{16}$$

Solution

Question 9

300%

600%

800%

80%

Solution

Question 10

$${ r }_{ n }={ a }_{ 0 }n-\beta $$

$${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }+\beta \quad $$

$${ r }_{ n }={ a }_{ 0 }{ n }^{ 2 }-\beta $$

$${ r }_{ n }={ a }_{ 0 }n+\beta $$

Solution

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