MCQ Questions for Class 9 Chemistry Structure Of The Atom Quiz 12

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

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$\beta$ -particles, which impinged on a metal foil and got absorbed
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$\gamma$ -rays, which impinged on a metal foil and ejected electrons
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helium atoms, which impinged on a metal foil and got scattered
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helium nuclei, which impinged on a metal foil and got scattered

Explanation

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

$D$

The radius of nucleus is:

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proportional to its mass number
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inversely proportional to its mass number
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proportional to the cube root of its mass number
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not related to its mass number

Explanation

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

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

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its atomic mass
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its atomic number
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the chemical nature of the atom
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number of the electron also

Rutherford's experiment established that:

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inside the atom there is a heavy positive centre
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nucleus contains protons and neutrons
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most of the space in an atom is empty
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size of nucleus is very small

Explanation

Based on Rutherford's experiment, he put forward the model of the atom, which had the following features:

$\left( 1 \right)$ There is a positively charged center in an atom called the nucleus nearly all the mass of an atom resides in the nucleus.

$\left( 2 \right)$ electrons revolve around the nucleus in well defined orbits

$\left( 3 \right)$ The size of the nucleus is very small when compared to that of the atom.

In the Bohr's model of the atom:

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the radius of $n$ th orbit is proportional to ${ n }^{ 2 }$
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the total energy of the electron in the $n$ th orbit is inversely proportional to ' $n$ '
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the angular momentum of the electron is integral multiple of ${ h }/{ 2\pi }$
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the magnitude of potential energy of an electron in an orbit is greater than kinetic energy

Explanation

According to the Bohr model of atom

Energy of electron in the

$n^{th}$ orbit of hydrogen like species

$E_n = -13.6 \cfrac{z^2}{n^2}eV$

Radius of

$n^{th}$ orbit of hydrogen like species

$R_n = R_0 \cfrac{n^2}{Z}$

Angular momentum of electron in the

$n^{th}$ orbit

$mvr = \cfrac{nh}{2\pi}$ integral multiple of

$\cfrac{h}{2\pi}$

Relation between energies

$-KE = TE= \cfrac{PE}{2}$

Comparing magnitudes =

$2|KE| = |PE|$

$\implies |PE| >|KE|$

The magnitude of potential energy is greater than that of kinetic energy.

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:

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Atomic number of element is 4.
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3 different types of spectral lines will be observed if electrons make transition upto ground state from the 2nd excited state.
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A 25 eV photon can set free the electron from the 2nd excited state of this sample
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2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atom

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

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Bohr
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Rutherford
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Moseley
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Thomson

Which of the following observations was not correct during Rutherford's scattering experiment?

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Most of the $\alpha$ -particles passed through the gold foil undeflected.
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A small fraction of the $\alpha$ -particles was deflected by small angles.
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A large number of the $\alpha$ -particles were bounced back.
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A very few $\alpha$ -particles (~1 in 12,000) were bounced back.

Explanation

Very few bombarded

$\alpha$ -particles on gold foil were deflected by sharp angles and therefore, it was concluded that the positive charge is concentrated in a very small volume of the atom called nucleus.

Most of the particles passed through the foil undeflected and was concluded that most of the space in an atom is empty.
A very few

$\alpha$ -particles (approx. 1 in 12,000) were bounced back.
So option

$C$ is only incorrect statement.

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?

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$-\dfrac {2.17\times 10^{-11}}{2}$
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$-\dfrac {2.17\times 10^{-11}}{2^{2}}$
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$-\dfrac {2.17\times 10^{-7}}{2^{2}}$
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$-\dfrac {2.17\times 10^{11}}{2^{2}}$

Explanation

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

$n=2$

$E_2=-\dfrac{2.178\times 10^{-11}}{2^2}ergatom^{-1}$

Which of the following conclusions regarding the structure of the atom is based on Rutherford's

$\alpha$ -particle scattering experiment?

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The positive charge is concentrated in a very small volume of the atom.
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The positive charge is scattered with the electrons throughout the atom.
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The volume occupied by the nucleus is half of the volume of atom.
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Most of the space in the atom is occupied by the neutrons.

Explanation

Very few bombarded

$\alpha$ -particles on gold foil were deflected by sharp angles and therefore, it was concluded that the positive charge is concentrated in a very small volume of the atom called a nucleus. Since few were deflected, the size of the nucleus is very small as compared with atom size.

option

$A$ is correct.

The density of nucleus is of the order of:

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${ 10 }^{ 5 }kg{ m }^{ -3 }$
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${ 10 }^{ 10 }kg{ m }^{ -3 }$
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${ 10 }^{ 17 }kg{ m }^{ -3 }$
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${ 10 }^{ 25 }kg{ m }^{ -3 }$

Explanation

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 $10^{17}\ Kg.m^{-3}$ .

Which of the following properties of atom could be explained correctly by Thomson's model of atom?

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Overall neutrality of atom
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Spectra of hydrogen atom
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Position of electrons, protons and neutrons in atom
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Stability of atom

Explanation

Hint: All atoms include tiny negatively charged subatomic particles or electrons, according to J.J. Thomson's work with cathode ray tubes.

Explanation:

According to the Thomson atom model:

1. An atom is spherical and has a radius of $10^{-10}$ meters.

2. The positive charge in the atom is evenly distributed.

3. The electrons are embedded in such a way that the atom's electrostatic configuration is the most stable.

As a result, the model was able to explain the atom's overall neutrality.

4. The atom's mass is considered to be evenly distributed across the atom.

It was unable to account for atomic spectra, atomic stability, or the real distribution of electrons, protons, and neutrons within the atom.

Final Answer:

The correct answer is option $(A)$ .

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?

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$2\rightarrow 5$
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$3\rightarrow 2$
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$5\rightarrow 2$
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$4\rightarrow 1$

Explanation

$\bf{Explanation-}$

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

$C$

Which of the following conclusions could not be derived from Rutherford's

$\alpha$ -particles scattering experiment?

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Most of the space in the atom is empty.
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The radius of the nucleus is about $10^5$ times less than the radius of the atom.
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Electrons move in a circular path of fixed energy called orbits.
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Electrons and the nucleus are held together by the electrostatic force of attraction.

Explanation

Based on Rutherford's

$\alpha$ -particles scattering experiment, the nuclear model of the atom was proposed as:

(i) Positive charge and most of the mass of the atom is densely concentrated in an extremely small region called the nucleus. The radius of the nucleus is about

$10^5$ times less than the radius of the atom.

(ii) The electrons revolve around the nucleus in circular paths.

(iii) Electrons and the nucleus are held together by electrostatic forces of attraction.

The revolution of the electron in a circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know. The concept of stationary state or orbits with fixed energy was given by Bohr's theory and not by Rutherford's theory.

In a Bohr's model of an atom when an electron jumps from n=1 to n=3, how much energy will be emitted or absorbed in erg?

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$2.15\times { 10 }^{ -11 }$
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$0.191\times { 10 }^{ -10 }$
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$2.389\times { 10 }^{ -12 }$
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$0.239\times { 10 }^{ -10 }$

Radius of

$2nd$ shell of

$He^+$ (where

$a_0$ - Bohr radius)

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$3a_0$
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$a_0$
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$\dfrac{3}{2}a_0$
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$2a_0$

Explanation

Radius of the Bohr's orbit is given by,

$r=\dfrac{h^2n^2}{4\pi^2 m Z e^2}$

Substituting values of h, m for helium ion and the electronic charge r may be calculated as

$r=\dfrac{0.529n^2}{Z}A^0$

Where

$0.529A^0$ is is the Bohr's radius

$a_0=0.529$

Substituting

$Z=2$ ,

$n=2$

$r=2a_0A^0$

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:

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$10^{-5}$
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$10^{5}$
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$10^{-15}$
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none of these

Explanation

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 =

$\dfrac{4}{3}\pi({10^{-8}})^3cm^3$

$\dfrac{V_N}{V_{atom}}$

$=$

$\dfrac{10^{-39}}{10^{-24}}$

$\dfrac{V_N}{V_{atom}}$

$=$

$10^{-15}$

How many sub-atomic particles are present in an

$\alpha$ - particle used in Rutherford's scattering experiment?

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Protons - 4, Neutrons - 0, Electrons - 0
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Protons - 2, Neutrons - 0, Electron - 2
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Protons - 2, Neutron - 2, Electron - 0
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Protons - 2, Neutron - 2, Electron - 1

Explanation

$\alpha$ 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

$\text{C}$ .

An atom can be smashed and broken into many nuclear particles by the high acceleration (increasing speed) imparted in:

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a chromatograph
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a cyclotron
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an electron microscope
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acathode-ray tube

What is the ratio of time periods

$(\dfrac{T_1}{T_2})$ in

$2^{nd}$ orbit of hydrogen atom to third orbit of

$He^+$ ion?

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$\dfrac{8}{27}$
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$\dfrac{32}{27}$
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$\dfrac{27}{32}$
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$\dfrac{27}{8}$

Explanation

$\dfrac{T_H}{T_{He}}=\dfrac{n_1^3t_2^2}{n_2^3t_1^2}$

$= \dfrac{8\times4}{3^3}=\dfrac{32}{27}$

The ratio is

$32:27$ .

A single electron orbits a stationary nucleus of charge

$+Z$ , where

$Z$ is a constant. It requires 47.2 eV to excite electron from second Bohr orbit to third Bohr orbit, find the value of

$Z$ .

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1
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3
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5
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4

Explanation

Position is

$n_1=2 \ \Rightarrow n_2=3$ and

$\Delta E=47.2eV$

We have

$\Delta E=13.6z^2(\dfrac{1}{n_1^2}− \dfrac{1}{n_2^2})eV$

$47.2=13.6z^2(\dfrac{1}{2^2}− \dfrac{1}{3^2})$

$z=5$

The nucleus of the atom

$\left( Z>1 \right)$ consists of:

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Proton and neutron
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Proton and electron
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Neutron and electron
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Proton, neutron and electrons

Explanation

The atomic nucleus is the central area of the atom. It is composed of two kinds of subatomic particles: protons and neutrons.The protons and neutrons held together to form the dense area of the nucleus.

For atoms havng

$Z>1$ nucleus contains protons and neutrons.

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____.

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$1.06 {\mathring A}$
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$0.265 {\mathring A}$
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$0.175 {\mathring A}$
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$0.53 {\mathring A}$

Explanation

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.

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

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$4.41\times {10}^{-16}J{atom}^{-1}$
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$-4.41\times {10}^{-17}J{atom}^{-1}$
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$-2.2\times {10}^{-15}J{atom}^{-1}$
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$-8.82\times {10}^{-17}J{atom}^{-1}$

Explanation

$IE$ 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

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

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$-13.6$ eV
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$-3.4$ eV
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$-1.5$ eV
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none of these

Explanation

$E_n = - 13.6\times(z^2/n^2)eV$

$E_3 = - 13.6\times1/9eV$

$= - 1.5eV$

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

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The electrons revolve around the nucleus in fixed energy orbits
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It is a stable model of atom which resembles our solar sysytem
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The distribution of electrons in the orbits was given by Rutherford
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It could not explain the atomic spectra of hydrogen

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:

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

Explanation

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.

Which one of the following is considered as the main postulate of Bohr's model of atom

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Protons are present in the nucleus
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Electrons are revolving around the nucleus
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Centrifugal force produced due to the revolving electrons balances the force of attraction between electron and protons
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Angular momentum of electron is an integral multiple of $\dfrac{h}{2\pi}$

Bohr's theory can be applied to determine the?

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electron gain enthalpy of hydrogen atom
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Electronegativity of ${ Li }^{ 2+ }$ ion
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second ionization energy of helium atom
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first ionization energy of Be atom

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

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$1 : 4$
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$1 : 8$
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$1 : 16$
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$16 : 1$

Explanation

Formula,

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

hydrogen,

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

$E=E_0$

$Be^{3+},$

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

$E=4E_0$

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

$1:4$

CBSE Questions for Class 9 Chemistry Structure Of The Atom Quiz 12 - MCQExams.com

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Practice Class 9 Chemistry Quiz Questions and Answers

CBSE Questions for Class 9 Chemistry Structure Of The Atom Quiz 12 - MCQExams.com

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Practice Class 9 Chemistry Quiz Questions and Answers

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