Atoms Test

Result

Atoms Test - 55

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The figure indicates the energy level diagram of an atoms and the origin of six spectrum lines in emission (e.g line no.$$5$$ arises from the transition from level $$B$$ to $$A$$) Which of the following spectral lines will occur in the absorption spectrum?

A
$$4,5,6$$

B
$$1,2,3,4,5,6$$

C
$$1.2.3$$

D
$$1,4,6$$

Solution

The absorption lines are obtained when the electron jumps from ground state $$(n = 1)$$ to the higher energy states. Thus only $$1, 2$$ and $$3$$ lines will be obtained.
Question 2
1 / -0

Find the frequency of revolution of the electron in the first stationary orbit of H-atom

A
$$8\times 10^{14}Hz$$

B
$$8.6\times 10^{10}Hz$$

C
$$8.6\times 10^{-10}Hz$$

D
$$8.12\times 10^{16}Hz$$

Solution

$$frame=\dfrac{1}{Time\ period}$$
Time perid$$=\dfrac{Total \ distance \ coverd}{velocity}=\dfrac{2\pi r}{v}$$
frequency$$=\dfrac{v}{2\pi r}$$
The velocity $$v_2$$ and radius $$r_2$$ of the $$2^{nd}$$ both orbit
$$ r_n=(0.53\times 10^{-10}n^2/Z)\ m$$ , $$Z=1\ and\ n=2 $$
$$r_2=0.53\times10^{-10}(2^2)\ =2.12\times 10^{-10}m$$
$$v_n=2.165\times 10^{6}\left(2/n\right)m/s$$
$$v_2=2.165\times 10^{6}\left(1/2 \right)=1.082\times 10^6m/s$$
frequency$$=\dfrac{v_2}{2\pi r_2}=\dfrac{1.082\times 10^6}{2(x)(2.12\times 10^{-10})}$$
$$f=8.13\times 10^{16}Hz$$
Question 3
1 / -0

In hydrogen atom, the electron is making $$6.6\times 10^{15}rev/s$$ around the nucleus in an orbit of radius $$0.528A^0$$. The equivalent magnetic dipole moment is :

A
$$1\times 10^{-15}Am^2$$

B
$$1\times 10^{-10}Am^2$$

C
$$1\times 10^{-23}Am^2$$

D
$$1\times 10^{-27}Am^2$$

Solution
Question 4
1 / -0

What will be relation between first bohr radius of H- atom and D-atom

A
$$r _ { H } = r _ { D }$$

B
$$r _ { H } > r _ { D }$$

C
$$r _ { \mathrm { H } } < \mathrm { r } _ { \mathrm { D } }$$

D
Can't predict

Solution
Question 5
1 / -0

The Bohr radius of the fifth electron of phosphorus (atomic number = 15) acting as dopant in silicon (relative dielectric constant = 12) is

A
10.6$$\ { A^0 }$$

B
0.53$$\ { { A ^0} }$$

C
21.2$$ { A^0 }$$

D
None of these

Solution

$${r_n} = { \in _r}\left( {\frac{{{n^2}}}{Z}} \right)$$
$${a_0} = 12 \times \frac{{{{\left( 5 \right)}^2}}}{{15}} \times 0.53 = 10.6\mathop A\limits^0 $$
Hence,
option $$(A)$$ is correct answer.
Question 6
1 / -0

The radius of the orbit of an electron in a Hydrogen-like atom is $$3{ a }_{ 0 }$$, where $${ a }_{ 0 }$$ is the Bohr radius. Its orbital angular momentum is $$\frac { 3h }{ 2\pi } $$. It is given that h is Planck's constant and R is Rydberg constant. The possible wavelength, when the atom de-excites is :

A
$$\dfrac { 4 }{ 5R } $$

B
$$\dfrac { 4 }{ 9R } $$

C
$$\dfrac { 1 }{ 2R } $$

D
$$\dfrac { 9 }{ 32R } $$

Solution

solution:
given ;
$$4.5a_0 = a_0 \dfrac{n^2}{Z}----(i)$$
$$\dfrac{nh}{2\pi} = \dfrac{3h}{2\pi} ----(ii)$$

so , n= 3 and z=2
so , possible wavelength are:

$$\dfrac{1}{\lambda _1}= RZ^2[\dfrac{1}{1^2} - \dfrac{1}{3^2}] \Rightarrow \lambda _1= \dfrac{9}{32R}$$

hence the correct option:D
Question 7
1 / -0

What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?

A
4 : 1

B
4 : 9

C
4 : 3

D
5 : 9

Solution

For transition $$n_2 \rightarrow n_1$$
the wavelength is given by $$\dfrac{1}{\lambda}=RZ^2(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$
For shortest wavelength in Lyman series put $$n_1=1$$ and $$n_2 \rightarrow \infty$$
and for shortest wavelength in Balmer series put $$n_1=2$$ and $$n_2 \rightarrow \infty$$
we get  $$\dfrac{1}{\lambda _L}=RZ^2$$
and  $$\dfrac{1}{\lambda _B}=RZ^2(\dfrac{1}{4})$$
ratio will be  $$\dfrac{\lambda _B}{\lambda _L}=\dfrac{4}{1}$$

Option A is correct.
Question 8
1 / -0

The number of bond pair of electrons and lone pair of electrons in $$O_2$$ molecules respectively are

A
2, 2

B
2 ,1

C
4, 2

D
2, 4

Solution
Question 9
1 / -0

In a cubical vessel $$1m \times 1m\times 1m$$ the gas molecules of diameter $$1.7 \times {10^{ - 8}}\;{\text{cm}}$$ are at a temperature 300 K and a pressure of $${10^{ - 4}}\;{\text{mm}}$$ mercury. The mean free path of the gas molecule is

A
1 meter

B
4 meter

C
2.42 meter

D
1 cm

Solution

The mean free path or average distance between collison for a gas molecule may be estimated from KTG(Kinetic Theory of Gases).
From Serway's approach:-
$$(Mean\quad free\quad path)\lambda=\cfrac{1}{\pi d^2n_v}$$
$$d\rightarrow $$ diameter of molecule
$$n_v\rightarrow$$ No. of molecules per unit volume
From idle gas law,
$$n_v=\cfrac{nN_A}{V}\Rightarrow n_v=\cfrac{N_AP}{RT}$$
So, from $$(i)$$
$$\lambda=\cfrac{RT}{\sqrt{2}\pi d^2N_A P}$$
A/Q we have
$$d=1.7\times 10^{-8}cm=17\times10^{-10}m\\P=10^{-4}mmHg\\T=300K\\ \lambda=\cfrac{8.314\times 300}{\sqrt2 \times3.14\times (1.7\times10^{-10})^2}(6.022\times10^{23}(10^{-4}))\\ \quad=\cfrac{8.314\times 3\times10^3}{1.414\times3.14\times2.89\times6.022}\\ \quad=2.4201m$$
Question 10
1 / -0

Suppose that the potential energy of a hypothetical atom consisting of a proton and an electron is given by U = - Ke$$^{2}$$ / 3 r$$^{3}$$.Then if Bhor's postulates are applied to this atom ,then the radius of the n $$^{th}$$ orbit will be proportional to:

A
n$$^{2}$$

B
1 / n$$^{2}$$

C
n$$^{3}$$

D
1 / n$$^{3}$$

Solution

$$u = \frac{{ - k{e^2}}}{{2{r^3}}}$$
$$F = \frac{{ - du}}{{dr}}$$
$$ = \frac{{k{e^2}}}{{{r^2}}}$$
$$\frac{{k{e^2}}}{{{r^2}}} = \frac{{m{v^2}}}{r}$$---------------$$(1)$$
$$mvr = \frac{{nh}}{{2\pi }}$$---------------$$(2)$$
$$ \Rightarrow r\alpha {n^2}$$
Hence,
option $$(A)$$ is correct answer.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Atoms Test - 55

Result

Atoms Test - 55

Score

-

Rank

-

SHARING IS CARING

Question 1

$$4,5,6$$

$$1,2,3,4,5,6$$

$$1.2.3$$

$$1,4,6$$

Solution

Question 2

$$8\times 10^{14}Hz$$

$$8.6\times 10^{10}Hz$$

$$8.6\times 10^{-10}Hz$$

$$8.12\times 10^{16}Hz$$

Solution

Question 3

$$1\times 10^{-15}Am^2$$

$$1\times 10^{-10}Am^2$$

$$1\times 10^{-23}Am^2$$

$$1\times 10^{-27}Am^2$$

Solution

Question 4

$$r _ { H } = r _ { D }$$

$$r _ { H } > r _ { D }$$

$$r _ { \mathrm { H } } < \mathrm { r } _ { \mathrm { D } }$$

Can't predict

Solution

Question 5

10.6$$\ { A^0 }$$

0.53$$\ { { A ^0} }$$

21.2$$ { A^0 }$$

None of these

Solution

Question 6

$$\dfrac { 4 }{ 5R } $$

$$\dfrac { 4 }{ 9R } $$

$$\dfrac { 1 }{ 2R } $$

$$\dfrac { 9 }{ 32R } $$

Solution

Question 7

4 : 1

4 : 9

4 : 3

5 : 9

Solution

Question 8

2, 2

2 ,1

4, 2

2, 4

Solution

Question 9

1 meter

4 meter

2.42 meter

1 cm

Solution

Question 10

n$$^{2}$$

1 / n$$^{2}$$

n$$^{3}$$

1 / n$$^{3}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.