Structure of Atom Test

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Structure of Atom Test - 51

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Question 1
1 / -0

Atomic radius is of the order of $$10^{-8}\ cm$$ and nuclear radius is of the order of $$10^{-13}\ cm$$. The fraction of atom occupied by nucleus is:

A
$$10^{-5}$$

B
$$10^{5}$$

C
$$10^{-15}$$

D
none of these

Solution

A nucleus is the positively charged center of the atom consisting of protons and neutrons. It's also known as the "atomic nucleus". Nearly all the mass of an atom is contained within the nucleus, since protons and neutrons have much more mass than electrons

according to qes.
Volume of nucleus = $$\dfrac{4}{3}{\pi}r^3$$ = $$\dfrac{4}{3}\pi({10^{-3}})^3 cm^3$$
Volume of atom = $$\dfrac{4}{3}\pi({10^{-8}})^3cm^3$$

$$\dfrac{V_N}{V_{atom}}$$ $$=$$ $$\dfrac{10^{-39}}{10^{-24}}$$

$$\dfrac{V_N}{V_{atom}}$$ $$=$$ $$10^{-15}$$
Question 2
1 / -0

If the diameter of two different nuclei are in the ratio 1 : 2, then their mass number are in the ratio:

A
1 : 2

B
8 : 1

C
1 : 8

D
1 : 4

Solution

Given:
Ratio of diameter of $$2$$ nuclei = $$1:2$$
To find the Ratio of their mass number.
As we know the - Radius of the nucleus is given as:
$$R=r_0A^{1/3}$$

$$A= Mass\ number$$, $$R=Radius\ of\ the\ nucleus$$, $$r_0=Nucleon\ number=1.25\ fm$$
As, $$\frac{D_1}{D_2}=\frac{1}{2}$$
Radius is just half of diameter as $$Radius\ \times \ 2=Diameter$$
So ratio of Radius$$=\frac{1}{2}$$

$$\frac{R_1}{R_2}=\frac{r_0\ A_1^{1/3}}{r_0\ A_2^{1/3}}$$
So, $$R^3\propto \ A$$
So Ratio of mass number i.e. $$\frac{A_1}{A_2}=\frac{1}{8}$$
optiin C is correct.
Question 3
1 / -0

Which series have highest energy in hydrogen spectrum?

A
Balmer

B
Bracket

C
Pfund

D
Lyman

Solution

Energy of electron varies with $$\dfrac{1}{n^2}$$ i.e principal quantum number. For lyman series principal quantum number$$(n)$$ is 1. Hence, Lyman series have highest energy in hydrogen spectrum.
Question 4
1 / -0

The Vividh Bharati station of all india Radio, Delhi,broadcasts on a frequency of 1,368kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to ?

A
Visible region

B
Radio wave region

C
Infrared region

D
Microwave region

Solution

$$F = \displaystyle \frac{c}{\lambda } \Rightarrow \lambda = \frac{c}{f} = \frac{3\times 10^{8}}{1.368\times 10^{3}} = 2.19 \times 10^{5} m $$
Hence, is lies in Radio-Wave Region.
Question 5
1 / -0

The electronic configuration, $$1s^22s^22p^63s^23p^63d^9$$, represents a:

A
metal atom

B
non-metal atom

C
non-metallic anion

D
metallic cation

Solution

The electronic configuration, $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^9}$$, represents a metallic cation because the $$e$$' of $$4s$$ Sulosshell are filled before $$3d$$ subshell.
Question 6
1 / -0

For the electronic transition from $$n=2$$ to $$n= 1$$ which one of the following will produce the shortest wavelength?

A
$$H$$ atom

B
$$D$$ atom

C
$${ He }^{ + }$$ion

D
$${ Li }^{ +2 }$$ion

Solution

According to Rydberg equation:
$$\dfrac{1}{\lambda} = R_Hz^2(1/n^2_1 -1/n^2_2)$$

As from the equation, the atomic number and wavelength are inversely related. More the atomic number, the shortest will be the wavelength.

The atomic number of lithium is 3 and is the highest of all. So, this nucleus will have the shortest wavelength.

Hence, option D is correct.
Question 7
1 / -0

In terms of Bohr radius $${a}_{0}$$, the radius of the second Bohr orbit of a hydrogen atom is given by:

A
$$8{a}_{0}$$

B
$$4{a}_{0}$$

C
$$2{a}_{0}$$

D
$$\sqrt {2}{a}_{0}$$

Solution

Since, $$r \propto {n^2}$$
Radius of $$2^{nd}$$ Bohr's orbit $$= 4a_0$$.
Question 8
1 / -0

If the ionisation energy of hydrogen is $$2.18\times { 10 }^{ -19 }\quad J$$, the first ionisation energy of $${ He }^{ + }$$ion is:

A
$$2.18\times { 10 }^{ -19 }\quad J$$

B
$$4.36\times { 10 }^{ -19 }\quad J$$

C
$$8.72\times { 10 }^{ -19 }\quad J$$

D
$$0.545\times { 10 }^{ -19 }\quad J$$

Solution

The energy of an electron in the first orbit $$= -2.18\times 10^{19} J$$
The energy of an electron in the fifth orbit

$$= \cfrac{-2.18\times 10^{-19}}{n_2}$$

$$= \cfrac{-2.18\times 10^{19}}{25}$$

$$E_5=8.72\times 10^{-19}$$ $$J$$

Hence, option $$C$$ is correct.
Question 9
1 / -0

A certain electronic transition from an excited state to the ground state of the $$H$$ atom in one or more steps gives rise to four lines in the ultraviolet region of the spectrum, how many lines does this transition produce in the infrared region of the spectrum.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let the four lines are observed in the ultraviolet region of the spectrum.

These correspond to the transitions

$$5 \to 1,4 \to 1,3 \to 1$$ and $$2 \to 1$$

The possible transitions in the IR region are

$$5 \to 4,5 \to 3$$ and $$4 \to 3$$

Therefore the infrared region of the spectrum, $$3$$ lines are produced.

Hence, te correct option is $$\text{C}$$
Question 10
1 / -0

Which of the following orbitals are in the increasing order of energy.

A
$$1s, 2s, 2p,3s$$

B
$$4s, 3s, 3p, 4d$$

C
$$5p, 4d, 5d, 4f, 6s$$

D
$$4p, 2s, 3s, 2p$$

Solution

Energy of the orbital $$\propto$$ Principal Quantum number
Energy of the orbital $$\propto$$ $$(n+l)$$ rule
If $$(n+l)$$ for two orbitals is same then the energy is judged by $$'n'$$. So order using this rule is and according to Aufbau's principle as shown in picture is:
$$1s<2s<2p<3s$$

Hence, option $$A$$ is correct.

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Structure of Atom Test - 51

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Structure of Atom Test - 51

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Question 1

$$10^{-5}$$

$$10^{5}$$

$$10^{-15}$$

none of these

Solution

Question 2

1 : 2

8 : 1

1 : 8

1 : 4

Solution

Question 3

Balmer

Bracket

Pfund

Lyman

Solution

Question 4

Visible region

Radio wave region

Infrared region

Microwave region

Solution

Question 5

metal atom

non-metal atom

non-metallic anion

metallic cation

Solution

Question 6

$$H$$ atom

$$D$$ atom

$${ He }^{ + }$$ion

$${ Li }^{ +2 }$$ion

Solution

Question 7

$$8{a}_{0}$$

$$4{a}_{0}$$

$$2{a}_{0}$$

$$\sqrt {2}{a}_{0}$$

Solution

Question 8

$$2.18\times { 10 }^{ -19 }\quad J$$

$$4.36\times { 10 }^{ -19 }\quad J$$

$$8.72\times { 10 }^{ -19 }\quad J$$

$$0.545\times { 10 }^{ -19 }\quad J$$

Solution

Question 9

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 10

$$1s, 2s, 2p,3s$$

$$4s, 3s, 3p, 4d$$

$$5p, 4d, 5d, 4f, 6s$$

$$4p, 2s, 3s, 2p$$

Solution

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