Structure of the Atoms Test

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Structure of the Atoms Test - 45

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Question 1
1 / -0

Rutherford's experiment on scattering of $$\alpha$$-particles showed for the first time that the atom has:

A
Electrons

B
Protons

C
Neutrons

D
Nucleus

Solution

Rutherford's experiment on scattering of α-particles showed for the first time that the atom has nucleus. He observed that the positively charged α-particles were repelled and deflected by the positive charges in the atom. Rutherford's named this positively charged portion of atom as nucleus.
Question 2
1 / -0

The introduction of a neutron into the nuclear composition of an atom would lead to a change in:

A
its atomic mass

B
its atomic number

C
the chemical nature of the atom

D
number of the electron also

Solution

The introduction of a neutron into the nuclear composition of an atom would lead to a change in its atomic mass.
However, its atomic number, the chemical nature of the atom and the number of the electron s will remain unchanged as they are related to the number of protons and are independent of the number of neutrons.

Note: The atomic mass is equal to the sum of the number of protons and the number of neutrons. Chemical properties depends on number of electrons which are equal to number of protons in neutral atom. They are also equal to atomic number.
Question 3
1 / -0

The nucleus of the atom consists of:

A
protons and neutrons

B
protons and electrons

C
neutrons and electrons

D
protons, neutrons and electrons

Solution

The nucleus of the atom consists of protons and neutrons. Electrons are present outside the nucleus.
Question 4
1 / -0

The mass of neutron is of the order of:

A
$${10}^{-27}$$ kg

B
$${10}^{-26}$$ kg

C
$${10}^{-25}$$ kg

D
$${10}^{-24}$$ kg

Solution

The mass of neutron is of the order of $$10^{-27}$$ kg. It is equal to $$1.674 \times {10}^{-27}$$ kg.
Question 5
1 / -0

The nucleus of an atom was discovered due to the experiment carried out by:

A
Bohr

B
Rutherford

C
Moseley

D
Thomson

Solution

The discovery of the nucleus of an atom was due to the experiment carried out by Rutherford. Rutherford's experiment on the scattering of α-particles showed for the first time that the atom has a nucleus. He observed that the positively charged α-particles were repelled and deflected by a positive charge in the atom. Rutherford's named this positively charged portion of the atom as the nucleus.
Question 6
1 / -0

The Ionisation potential of Hydrogen is $$2.17\times 10^{-11}erg/ atom$$. The energy of the electron in the second orbit of the hydrogen atom is?

A
$$-\dfrac {2.17\times 10^{-11}}{2}$$

B
$$-\dfrac {2.17\times 10^{-11}}{2^{2}}$$

C
$$-\dfrac {2.17\times 10^{-7}}{2^{2}}$$

D
$$-\dfrac {2.17\times 10^{11}}{2^{2}}$$

Solution

Energy of electron hydrogen atom is given by,
$$E_n=-\dfrac{2.178\times 10^{-11}}{n^2}ergatom^{-1}$$
Therefore, for second orbit i.e $$n=2$$
$$E_2=-\dfrac{2.178\times 10^{-11}}{2^2}ergatom^{-1}$$
Question 7
1 / -0

Rutherford's experiment, which established the nuclear model of the atom, used a beam of:

A
$$\beta$$-particles, which impinged on a metal foil and got absorbed

B
$$\gamma$$-rays, which impinged on a metal foil and ejected electrons

C
helium atoms, which impinged on a metal foil and got scattered

D
helium nuclei, which impinged on a metal foil and got scattered

Solution

Rutherford's experiment, which established the nuclear model of the atom, used a beam of helium nuclei ( $$\displaystyle \alpha -$$particles), which impinged on a gold foil ( of thickness 100 nm) and got scattered. The gold foil was surrounded by $$ZnS$$ fluorescent screen. Wherever alpha particles strike the screen, a tiny flash of light was produced at that point.

$$\alpha-$$ particles are called a helium nucleus.

Hence, the correct option is $$D$$
Question 8
1 / -0

The density of nucleus is of the order of:

A
$${ 10 }^{ 5 }kg{ m }^{ -3 }$$

B
$${ 10 }^{ 10 }kg{ m }^{ -3 }$$

C
$${ 10 }^{ 17 }kg{ m }^{ -3 }$$

D
$${ 10 }^{ 25 }kg{ m }^{ -3 }$$

Solution

Correct Answer: Option (C).

Hint: Nucleus contains protons and neutrons.

Explanation:
Nucleus is positively charged having positively charged proton and neutral neutron. It is the central part of an atom.
The atomic nucleus is a largely dense region in the center of the atom.
The nuclear force is a natural force of attraction that holds protons and neutrons together

Nucleus has density $$2.3\times 10^{17}\ Kg/m^3$$.

So, its density is of the order of $$10^{17}\ Kg.m^{-3}$$.

Final Answer: The density of the nucleus is of the order of $$10^{17}\ Kg.m^{-3}$$.
Question 9
1 / -0

The radius of nucleus is:

A
proportional to its mass number

B
inversely proportional to its mass number

C
proportional to the cube root of its mass number

D
not related to its mass number

Solution

The radius of a nucleus (r) is proportional to the cube root of its mass number (A).
The approximate law is $$r = r_0 \times A^{ 1/3 }$$
where $$r_0 = 1.2 \times 10^{ -15 }\ m$$
option C is correct
Question 10
1 / -0

In Bohr series of lines of hydrogen spectrum, the third line from the red corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

A
$$2\rightarrow 5$$

B
$$3\rightarrow 2$$

C
$$5\rightarrow 2$$

D
$$4\rightarrow 1$$

Solution

$$\bf{Explanation-}$$
Line Spectrum of Hydrogen like atoms are given by:
$$\dfrac{1}{\lambda} = RZ^2(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2})$$
Where R is called Rhydberg constant, $$R = 1.097 \times 10^7$$,
Z is atomic number.
$$n_1= 1,2 ,3….$$
$$n_2= n_1+1, n_1+2 ,……$$

The electron has minimum energy in the first orbit and its energy increases as $$n$$ increases. Here $$n$$ represents the number of orbits.

Red line means it is the visible spectra and visible spectra means it is Balmer series that is $$n=2$$
So, the third line after $$n=2$$ will be $$n=5$$ that is Pfund series.
Therefore, the electrons jump from $$n=5$$ to $$n=2$$.

$$\bf{Correct \ Option}-$$ Option $$C$$

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Structure of the Atoms Test - 45

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Structure of the Atoms Test - 45

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Question 1

Electrons

Protons

Neutrons

Nucleus

Solution

Question 2

its atomic mass

its atomic number

the chemical nature of the atom

number of the electron also

Solution

Question 3

protons and neutrons

protons and electrons

neutrons and electrons

protons, neutrons and electrons

Solution

Question 4

$${10}^{-27}$$ kg

$${10}^{-26}$$ kg

$${10}^{-25}$$ kg

$${10}^{-24}$$ kg

Solution

Question 5

Bohr

Rutherford

Moseley

Thomson

Solution

Question 6

$$-\dfrac {2.17\times 10^{-11}}{2}$$

$$-\dfrac {2.17\times 10^{-11}}{2^{2}}$$

$$-\dfrac {2.17\times 10^{-7}}{2^{2}}$$

$$-\dfrac {2.17\times 10^{11}}{2^{2}}$$

Solution

Question 7

$$\beta$$-particles, which impinged on a metal foil and got absorbed

$$\gamma$$-rays, which impinged on a metal foil and ejected electrons

helium atoms, which impinged on a metal foil and got scattered

helium nuclei, which impinged on a metal foil and got scattered

Solution

Question 8

$${ 10 }^{ 5 }kg{ m }^{ -3 }$$

$${ 10 }^{ 10 }kg{ m }^{ -3 }$$

$${ 10 }^{ 17 }kg{ m }^{ -3 }$$

$${ 10 }^{ 25 }kg{ m }^{ -3 }$$

Solution

Question 9

proportional to its mass number

inversely proportional to its mass number

proportional to the cube root of its mass number

not related to its mass number

Solution

Question 10

$$2\rightarrow 5$$

$$3\rightarrow 2$$

$$5\rightarrow 2$$

$$4\rightarrow 1$$

Solution

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