Structure of the Atoms Test

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Structure of the Atoms Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

How many sub-atomic particles are present in an $$\alpha $$- particle used in Rutherford's scattering experiment?

A
Protons - 4, Neutrons - 0, Electrons - 0

B
Protons - 2, Neutrons - 0, Electron - 2

C
Protons - 2, Neutron - 2, Electron - 0

D
Protons - 2, Neutron - 2, Electron - 1

Solution

$$\alpha$$ particle is identical to a helium atom that has been lost its two electrons.
no. of $$e^{-}=0$$
no. of proton $$=2$$
no. of neutron $$=2$$
Hence, the correct option is $$\text{C}$$.
Question 2
1 / -0

Energy of the third orbit of Bohr's atom is:

A
$$-13.6$$ eV

B
$$-3.4$$ eV

C
$$-1.5$$ eV

D
none of these

Solution

$$E_n = - 13.6\times(z^2/n^2)eV$$
$$E_3 = - 13.6\times1/9eV$$
$$ = - 1.5eV$$
Question 3
1 / -0

Which of the following statements is incorrect regarding Rutherford's model of atom?

A
The electrons revolve around the nucleus in fixed energy orbits

B
It is a stable model of atom which resembles our solar sysytem

C
The distribution of electrons in the orbits was given by Rutherford

D
It could not explain the atomic spectra of hydrogen

Solution

Postulates of Rutherford nuclear model:
The positive charge is concentrated in the center of the atom, called nucleus.
Electrons revolve around the nucleus in circular paths called orbits.
The nucleus is much smaller in size than the atom.
It is a stable model of atom that resembles our solar system.

Atomic spectra of hydrogen were proposed by Bohr.
The distribution of electrons in the orbits was not given by Rutherford. It was given by Bohr and Bury.
Question 4
1 / -0

The radius of $$He^+$$ ion is $$x\ \mathring{A}$$ in its ground state. The radius of $$Li^{2+}$$ ion in the ground state ($$ \mathring{A}$$ ) is:

A
$$\dfrac{2x}{3}$$

B
$$\dfrac{3x}{2}$$

C
$$\dfrac{2}{3x}$$

D
$$\dfrac{3}{2x}$$

Solution

Radius of an orbit $$= 0.529\times \dfrac {n^2}{Z}$$

$$R \propto \dfrac 1Z$$

$$\dfrac{R_1}{R_2} = \dfrac{Z_2}{Z_1}$$

$$\dfrac {x}{R_2} = \dfrac 32$$

$$R_2 = \dfrac{2x}{3}$$

Hence, option A is correct.
Question 5
1 / -0

Bohr's theory can be applied to determine the?

A
electron gain enthalpy of hydrogen atom

B
Electronegativity of $${ Li }^{ 2+ }$$ ion

C
second ionization energy of helium atom

D
first ionization energy of Be atom
Question 6
1 / -0

The radius of the hydrogen atom in the ground state is $$0.53 {\mathring A}$$, the radius of $$L{i^{2 + }}$$ in a similar state is____.

A
$$1.06 {\mathring A}$$

B
$$0.265 {\mathring A}$$

C
$$0.175 {\mathring A}$$

D
$$0.53 {\mathring A}$$

Solution

Radius of hydrogen like particle $$r_n=a_0\dfrac{n^2}{2}$$
$$0.530 \mathring { A }=a_0\dfrac{1^2}{1}$$
$$a_0=0.53\mathring{A}$$
for $$Li^{2+}$$ $$r=a_0\dfrac{1^2}{3}$$
$$=\dfrac{0.530}{3}=0.176 \mathring{A}$$
Hence, option $$C$$ is correct.
Question 7
1 / -0

Ionization energy of $${He}^{+}$$ is $$19.6\times {10}^{-18}J{atom}^{-1}$$. The energy of the first stationary state ($$n=1$$) of $${Li}^{2+}$$ is

A
$$4.41\times {10}^{-16}J{atom}^{-1}$$

B
$$-4.41\times {10}^{-17}J{atom}^{-1}$$

C
$$-2.2\times {10}^{-15}J{atom}^{-1}$$

D
$$-8.82\times {10}^{-17}J{atom}^{-1}$$

Solution

$$ IE$$ of $$ { He }^{ + }$$=$$19.6\times { 10 }^{ -18 } J$$ per atom
$$ IE\quad =\quad { E }_{ 1\quad }(for\quad H)\times { Z }^{ 2 }$$
$${ E }_{ 1 }\times 4= -19.6\times { 10 }^{ -18 }\quad J$$
$${ E }_{ 1 }(for \; {Li}^{+2})={ E }_{ 1 }(for\quad H)\times 9$$
=$$- \cfrac { 19.6\times { 10 }^{ -18 } }{ 4 } \times 9$$
=$$- 441\times { 10 }^{ -18 }$$
=$$- 4.41\times { 10 }^{ -17 } J$$per atom
Question 8
1 / -0

The ratio of the energy of the electron in the ground state of hydrogen to the electron in first excited state of $$Be^{3+}$$ is

A
$$1 : 4$$

B
$$1 : 8$$

C
$$1 : 16$$

D
$$16 : 1$$

Solution

Formula,

$$E=E_0 \dfrac{z^2}{n^2}$$

hydrogen,

$$E=E_0 \dfrac{z^2}{n^2}=E_0 \dfrac{1^2}{1^2}$$

$$E=E_0$$

$$Be^{3+},$$

$$E=E_0 \dfrac{z^2}{n^2}=E_0 \dfrac{4^2}{2^2}$$

$$E=4E_0$$

$$Ratio=\dfrac{E_0}{4E_0}$$

$$1:4$$
'
Question 9
1 / -0

Valency expresses:

A
total electrons in an atom

B
atomicity of an element

C
oxidation number of an element

D
combining capacity of an element

Solution

Valency is the number of electrons lost, gained, or shared by an atom during a chemical reaction. Elements in the same group have the same valency because they have the same number of electrons in the outermost shell.
Therefore, valency is combining the capacity of an element.
Hence, the correct option is $$D$$.
Question 10
1 / -0

In a hydrogen like sample electron is in 2nd excited state,the energy of the 4th state of this sample is -13.6 eV,then the incorrect statement is:

A
Atomic number of element is 4.

B
3 different types of spectral lines will be observed if electrons make transition upto ground state from the 2nd excited state.

C
A 25 eV photon can set free the electron from the 2nd excited state of this sample

D
2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atom

Solution

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Structure of the Atoms Test - 47

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Structure of the Atoms Test - 47

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Question 1

Protons - 4, Neutrons - 0, Electrons - 0

Protons - 2, Neutrons - 0, Electron - 2

Protons - 2, Neutron - 2, Electron - 0

Protons - 2, Neutron - 2, Electron - 1

Solution

Question 2

$$-13.6$$ eV

$$-3.4$$ eV

$$-1.5$$ eV

none of these

Solution

Question 3

The electrons revolve around the nucleus in fixed energy orbits

It is a stable model of atom which resembles our solar sysytem

The distribution of electrons in the orbits was given by Rutherford

It could not explain the atomic spectra of hydrogen

Solution

Question 4

$$\dfrac{2x}{3}$$

$$\dfrac{3x}{2}$$

$$\dfrac{2}{3x}$$

$$\dfrac{3}{2x}$$

Solution

Question 5

electron gain enthalpy of hydrogen atom

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

second ionization energy of helium atom

first ionization energy of Be atom

Question 6

$$1.06 {\mathring A}$$

$$0.265 {\mathring A}$$

$$0.175 {\mathring A}$$

$$0.53 {\mathring A}$$

Solution

Question 7

$$4.41\times {10}^{-16}J{atom}^{-1}$$

$$-4.41\times {10}^{-17}J{atom}^{-1}$$

$$-2.2\times {10}^{-15}J{atom}^{-1}$$

$$-8.82\times {10}^{-17}J{atom}^{-1}$$

Solution

Question 8

$$1 : 4$$

$$1 : 8$$

$$1 : 16$$

$$16 : 1$$

Solution

Question 9

total electrons in an atom

atomicity of an element

oxidation number of an element

combining capacity of an element

Solution

Question 10

Atomic number of element is 4.

3 different types of spectral lines will be observed if electrons make transition upto ground state from the 2nd excited state.

A 25 eV photon can set free the electron from the 2nd excited state of this sample

2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atom

Solution

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