Unit 02 Structure Of Atom

The term atom was introduced by Dalton and is defined as the smallest particle of an element that retains all its properties and identity during a chemical reaction.

Thomson’s Model of Atom :

According to Thomson an atom is a sphere of positive charge in which the small negatively charged particles (electrons) are embedded. The number of electrons is sufficient to neutralize the positive charge.

There are three fundamental particles which constitute an atom. These are named electrons, protons and neutrons.

Rutherford’s Model of atom :

This model was based upon the results of the famous alpha-rays scattering experiment. It is also known as ‘Planetary model’ and can be summarised in following points.

(i) Most of the mass and all the positive charge of an atom are present in a very small region called the nucleus.

(ii) The magnitude of the charge on the nucleus is different in atoms of different elements.

(iii) Electrons revolve in the space around nucleus in different circular orbits and the number of electrons is equal to the number of units of positive charge in the nucleus.

Atomic number (Z)

Number of protons in the nucleus of an atom = Number of electrons in the extra nuclear part of the electrically neutral atom.

Mass number (A)

Number of protons + Number of neutrons in the nucleus of an atom.

Types of atomic species

Isotopes : Atoms of same element having same atomic number but different mass number. For example, isotopes of Hydrogen are 1H,12H,1H

Isobars: Atoms of different elements having the same mass number but different atomic number, e.g., 4018Ar,4019 K40,20Ca

Isotones: Atoms of different elements which contain the same number of neutrons. e.g., 146C, 157 N,1680

Isoelectronic Species : Atoms or ions containing the same number of electrons.

For example, N3,O2,F,Na+,Mg2+,Al3+ and Ne, having 10 electrons each are isoelectronic

Dual nature of light

Some properties of light can be explained only by considering the wave nature (differaction and interference) while some could be explained by the particle nature (photoelectric effect).

Characteristics of a wave

Wavelength (λ). It is the distance between any two consecutive crests or troughs. It is expressed in \AA,m,cm, pm or nm.. (1\AA=1010 m or 108 cm,1 nm=109 m,1pm=1012 m).

Frequency (v). It is equal to number of waves passing through a point in one second. Units are hertz or s1(1 Hz=1 s1).

Velocity (c). It is the distance travelled by the wave in one second.

Relationship between c,v and λ:c=v×λ.

Amplitude (a). It is the height of the crest or depth of the trough.

Wave number (v¯). It is equal to reciprocal of wavelength (v¯=1λ).

Electromagnetic spectrum. It is the arrangement of electromagnetic radiations in order of increasing wavelengths :

Cosmic rays <γ-rays <X-rays <UV<visible<IR<Microwaves< Radiowaves

Planck’s quantum theory

Radiant energy is emitted or absorbed discontinuously in the form of small packets of energy called quanta.

Energy of each quantum e=hv, h= Planck’s constant =6.626×1034Js

Total energy emitted or absorbed =nhv ( n is an integer)

Photoelectric effect:

When radiation of certain minimum frequency (v0) strikes the surface of a metal, electrons are ejected.

The photoelectric effect was finally explained by Albert Einstein in the year 1905 with the help of Planck’s quantum theory. According to this theory, the energy of one quantum of radiation depends upon its frequency. Electron would be ejected only if the energy possessed by one quantum is at least sufficient to overcome the force by which the electron is held. Thus, it must have certain minimum energy which is called threshold energy or work function. Any excess energy given by the quantum would go as kinetic energy of the ejected electron. Since according to Planck’s theory, the energy possessed by a quantum is directly proportional to the frequency of the radiation used, the kinetic energy of the ejected electron is also proportional to the frequency.

According to Planck’s Theory

E=hv

If threshold frequency for a metal is v0, the energy required to make the electron just free is hv0. The kinetic energy of the ejected electron would be equal to the difference is the energy of one quantum of the radiation used and the binding energy of the electron

ΔE=hvhv0

Where ΔE represents the kinetic energy of the ejected electron which can be equated with it’s velocity

ΔE=1/2mv2

or it can be written as

1/2mv2=hvhv0

This concept could not be explained by the classical wave theory but the quantum theory could provide a satisfactory explanation.

Emission spectrum of Hydrogen atom

A spectrum having distinct lines was observed for hydrogen. Rydberg gave a general relation between wave number and series of integers

1λ=v¯=109677(1n121n22)cm1

Where 109677 cm1 is called the Rydberg constant.

Depending upon the different valves of n1 and n2, these spectral lines are devided into following groups (also called series). For each group of lines n1 is constant and n2 varies and have values (n1+1) and higher.

Series Region n1 n2
Lyman UV 1 2,3,
Balmer Visible 2 3,4,
Paschen IR 3 4,5,
Brackett IR 4 5,6,
Pfund IR 5 6,7,..
Humphreys Far IR 6 7,8,..

These spectral lines split up further in very closely spaced lines when the sample is placed in a magnetic field. This phenomenon is known as Zeeman effect.

Bohr’s Model of Atom

Niel Bohr modified Rutherford’s model by adding the concept of quantization of energy. He conceptualised that the electrons revolve around the nucleus on fixed circular paths called orbits without losing or gaining energy. These are also called energy levels or shells. The energy of different orbits in an atom in given by the expression

En=2π2me4n2h2

For hydrogen like species (i.e. having one electron only) this expression is written as

En=2π2z2me4n2h2

Substituting the values of m (mass of electron), e (charge on electron) and h (Planck’s constant) in the expression for hydrogen atom we get

En=2.178×1018n2Jatom1 or  En=13.595n2eVatom1 or  En=1312n2 kJ mol1

Radius of a Bohr orbit is given as

rn=n2h24π2Zme2 rn=5.29×1011n2Zm

For the first orbit of hydrogen

r=5.29×1011 m =52.9pm

The change in energy when an electron goes from one orbit to another is expressed as

ΔE=Einal Einital 

The energy gain or loss takes place by the absorption or emission of radiation which are related as

ΔE=hv=hcλ=hcv¯

Limitations of Bohr’s Model

It could notexplain:

Line spectra of multi electron atoms

Splitting of lines in magnetic field (Zeeman effect) and in electric field (Stark effect).

Three dimensional model of atom

Significance of Negative sign in energy expression

The negative sign in the energy expression appears because the energy of a free electron at rest is taken as zero. Such an electron would be at infinite distance away from the nucleus and would not experience any force of attraction towards it. As it comes closer, it experiences stronger and stronger attractive force and its energy would decrease. Since it is less than zero, it would have negative sign. The energy of the electron would be lowest in the orbit with n=1 and this most stable state is known as the ground level. If the electron has to jump to the next orbit with n=2, some energy must be given to it to overcome the attractive force of the nucleus. Thus in the second orbit, the energy of the electron is greater than is the first. Similarly, as the electron shifts from the second to the third, fourth and so as, its energy would increase.

Dual nature of matter and de Broglie Equation

The French physicist Louis de Broglie postulated the dual nature of moving particles and showed that wavelength (λ) of the matter wave is related to the momentum ( p ) by the equation

λ=hp=hmv

Where m= mass, v= velocity and h is Planck’s constant.

Heisenberg’s uncertainity principle - It is impossible to measure simultaneously the position and momentum of a microscopic particle with absolute accuracy. If one of them is measured with great accuracy, the other becomes less accurate. Δx.Δph4π,Δx= uncertainity in position

Δp= uncertainity in momentum

Δx.mΔVh4π,ΔV= uncertainity in velocity

Significant only for microscopic particles

Quantum Mechanics

Quantum mechanics takes into account the dual behaviour of matter. An equation, given by Schrodinger, which has a better physical interpretation in terms of wave properties is

H^ψ=Eψ

Where H^ is called hamiltonian and E is the total energy of the electron and ψ is a mathematical function, called wave function. Its square ψ2 is proportional to the probability of finding the electron at a given point. For this reason ψ is also called probability amplitude. On solving the schrodinger wave equation, three integers appear which can have only some restricted values. These integers are called quantum numbers. The three quantum numbers thus obtained are (i) the principal quantum number ( n), (ii) the azimuthal or subsidiary quantum number (I) and (iii) the magnetic quantum number (m1). A fourth quantum number is required to describe the electronic configuration of different elements which is called the spin quantum number (ms).

Significance of Quantum Numbers

Principal Quantum Number ( n ) identified the main energy level to which the electron belongs. It also specifies the average distance of the electron from the nucleus. It can have any integral value greater than zero i.e., n=1,2,3, The various value of n are sometimes also designated by the letters K,L,M, N,

Azimuthal Quantum Number or Angular Momentum Quantum Number (I) determines the shape of an orbital and the angular momentum of an electron occupying that orbital. It can have values from 0 to (n1), each of which represents a different sub-energy level or sub-shell. These sub-shells are designated as s,p,d,f according to the value of I=0,1,2,3 respectively. The shapes of s,p and d subshells are shown below.

Magnetic Quantum Number ( m, or m ) determines the preferred orientations of orbitals in space. The permitted values for m varies from - I to + lincluding zero.

Spin Quantum Number ( s or ms ) arises due to the spinning of the electron about its own axis. The spin can be clockwise represented by +1/2 or anti-clockwise represented by 1/2.

Electronic Configuration of Atoms

The arrangement of electrons in various orbitals is called the electronic configuration. This arrangement is obtained on the basis of following rules, which constitute the Aufbau Principle.

(i) The minimum energy rule

A sub-shell (group of orbitals) with lower energy is filled up first before filling of the subshell with higher energy begins. In other words, the electron must occupy the subshell of the lowest energy. The order of energy can be remembered as follows :

(a) Lower the value of (n+1), lower is the energy of the subshell. For one electron systems like H,He+,Li2+ etc., the energy of electrons is governed only by the value of n and not by (n+l).

(b) If the two subshells have the some (n+l) value, the one with lower n value will have the lower energy. Following these points, the sequence of orbitals for filling of electrons comes out to be :

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s,5f,6d,7p and so on.

(ii) Pauli’s Exclusion Principle

It states that no two electrons in an atom can have the same set of values of all the four quantum numbers. It can also be said that in an orbital, there can be maximum of two electrons and they should be of opposite spin.

(iii) Hund’s Rule of Maximum Multiplicity

According to this rule electron pairing in any of the p,d or f orbital does not occur until all the orbitals of that sub-shell are singly occupied.

Suppose in an atom, two electrons have to occupy a p-subshell which has three orbitals. px,py and pz all of which are of identical energy, In such a situation, two arrangements are possible. Both the electrons may occupy the same orbital or different orbitals. The two arrangements may be represented as px2,py0, pz0 or px1,py1,pz0. In the first arrangement, the electrons are paired while in the second, they are unpaired. According to Hund’s rule, the more stable arrangement of the two is px1,py1,pz in which both the electrons are unpaired. Thus, if a p-subshell has three electrons, the more stable arrangement would be px1,py1,pz1 in which all the three electrons are unpaired.

Half-filled and completely filled sub-shells are more stable due to symmetry or we can say due to exchange energy. Hence the configurations ( n1)d5ns1 and (n1)d10ns1 are more stable than ( n1)d4ns2 and (n1)d9ns2 respectively. This is reflected in the configuration of elements like Cr(z=24) and Cu(z= 29), whose configurations are 1s22s22p63s23p63d54s1 and 1s22s22p63s23p63d104s1, respectively.

Solved Problems

Question 1. What is the energy of an electron (m=9.1×103 kg) moving with a speed 5.0×107 kilometre per second?

(a) 11.4×1016 J

(b) 11.4×1010 J

(c) 22.75×1024 J

(d) 22.75×1021 J

Show Answer

Answer (b)

E=1/2mvv2

Given v=5.0×107kms1=5.0×1010 ms1

Substituting the value

E=12×9.1×103 kg×(5.0×1010 ms1)2

=11.4×1010kgm2 s2

=11.4×1010 J

Question 2. What is the momentum of a particle which has a wave length of 2A ?

(a) 3.3×1024 kg ms1

(b) 3.3×1034 kg ms1

(c) 3.3×1026 kg ms1

(d) 13.2×1044 kg ms1

Show Answer

Answer (a)

Momentum p=hλ

h=6.6×1034kgm2 s1

Given λ=2\AA=2×1010 m

Substituting the values

p=6.6×1034kgm2 s12×1010 m

=3.3×1024kgms1

Question 3. What is the ratio of energies of an electron is 1 st orbits of He+and H ?

(a) 1:1

(b) 1:2

(c) 2:1

(d) 4:1

Show Answer

Answer (d)

We know that

En=1312Z2n2 kJ mol1

For H,Z=1 and n=1

EH=1312(1)2=1312 kJ mol1

For He+Z=2 and n=1

EHe=1312(2)2=5248 kJ mol1 EHeEH=5248 kJ mol11312 kJ mol1=4

Question 4. Radius of which orbit of He+is one half as that of 4 th orbit of Be3+ ?

(a) 1

(b) 2

(c) 6

(d) 8

Show Answer

Answer (b)

Rn =5.29×1011×n2Z m

For He+:Z=2,n= ?

FeBe3+:Z=4 and n=4

(rn)He+=1/2(r4)Be3+

5.29×1011xn22=12×5.29×1011×(4)24

n2=4

n=2

So 2 nd orbit of He+would have one half radius than that of 4 th orbit of Be3+.

Question 5. What are the possible values of four enantum numbers (n,l,m,s) for an electron in 4f orbital.

(a) 4,2,2,+1/2

(b) 4,3,2,+1/2

(c) 5,3,2,1/2

(d) 3,3,1,+12

Show Answer

Answer (b)

For an electron is 4f orbital

n=4 and for ’ f ’ orbital I=3

Possible values for m=3,2,1,0,+1,+2,+3

and for S=+1/2 and 1/2

So out of given choices ‘b’ is the correct choice.

Question 6. What is the uncertainty in the position of a particle if the uncertainty is its momentum is 3.3×102 kgms1.

(a) 1.6×1035 m

(b) 3.2×1033 m

(c) 3.2×1034 m

(d) 1.6×1033 m

Show Answer

Answer (d)

We know that

ΔxΔph4π

or Δx=h4πΔp

Where =6.62×1034 kg m2 s1

and π=3.1416

Substituting the values

Δx=6.62×1034kgm2 s14×3.1416×3.3×102kgms1

=1.6×1033 m

Question 7. If the energy difference between the ground state of an atom and its excited state is 3.0×1019 J, what is the wavelength of the photon required to produce this transition?

(a) 6.62×109 cm

(b) 6.62×105 m

(c) 6.62×107 m

(d) 6.62×107 cm

Show Answer

Answer (c)

λ=hcΔE

Substituting the values h=6.62×1034Js

c=3×108 ms1

and ΔE=3×1019 J

λ=(6.62×1034Js)(3.0×108 ms1)3×1019 J

=6.62×107 m

Question 8. The ionisation energy of H-atom (in the ground state) is kJ. The energy required for an electron to jump from 2nd to 3rd orbit will be

(a) x/6

(b) 5x

(c) 7.2x

(d) 5x/36

Show Answer

Answer (d)

The ionisation energy in ground state is kJ,

so the energy of 1st orbit (E1)=xkJ

Energy of 2nd orbit (E2)=x(2)2=x4kJ

Energy of 3rd orbit (E3)=x(3)2=x9 kJ

Energy required to jump from 2nd to 3rd orbit

is E3E2=X9(X4)

=x4x9=5x36

Question 9. What is the de Broglie wavelength of a 66 kg man skiing down Kufri hill in Simla at 1×103 msec1 ? (Planck’s constant =6.6×1034Js )

(a) 1×1036 m

(b) 1×1037 m

(c) 1×1038 m

(d) 1×1039 m

Show Answer

Answer (c)

λ=hp=hmv

Given That

h=6.6×1034 kg m2 s1

m=66 kg

v=1×103 msec1

Substituting the values

λ=6.6×103466×103=1×1038 m

Question 10. Which of the following has more unpaired d-electrons?

(a) Fe2+

(b) Zn+

(c) Ni3+

(d) Cu+

Show Answer

Answer (a)

The electronic configurations for various species are

Zn+:1s22s22p63s23p64s13d10

Fe2+:1 s22 s22p63 s23p63 d6

Ni3+:1 s22 s22p63 s23p63 d7

Cu+:1s22s22p63s23p63d10

Thus maximum number of unpaired electrons are present in Fe2+

Question 11. Which of the following sets of quantum numbers represents the highest energy level in an atom?

(a) n=3,I=1,m=1,s=+1/2

(b) n=3,I=2,m=1,s=+1/2

(c) n=4,I=0,m=0,s=+1/2

(d) n=3,I=0,m=0,s=+1/2

Show Answer

Answer (b)

According to (n+I) rule, the energy of an orbital depends upon the sum of the values of the principal quantum number (n) and the azimuthal quantum number (I). Lower the value of (n+I), lower is the energy. If two different orbitals have the same value of (n+I), the orbital with lower value of n has lower energy.

In option (a), (n+I)=3+1=4

In option (b),(n+l)=3+2=5

In option (c),(n+l)=4+0=4

In option (d),(n+I)=3+0=3

Thus highest energy level is depicted in option (b).

Question 12. The ionisation enthalpy of hydrogen atom is 1.312×106 J mol. The energy required to excite the electron in the atom from n1=1 to n2=2 is

(a) 8.51×105 J mol

(b) 7.56×105 J mol

(c) 6.56×105 J mol

(d) 9.84×105 J mol

Show Answer

Answer (d)

Ionisation enthalpy of hydrogen atom is 1.312×106 J mol. This suggests that the energy of electron in the ground state (first orbit) is 1.312×106 J mol.

ΔE=E2E1

=(1.312×106(2)2)(1.312×1061)

=9.84×105 J mol1

Correct option is (d)

Question 13. In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005. Certainty with which the position of an electron can be located is (h=6.6×1034 kg m2 s1; mass of electron em=9.1×1031 kg )

(a) 1.52×104 m

(b) 5.10×103 m

(c) 1.92×103 m

(d) 3.84×103 m

Show Answer

Answer (c)

According to Heisenberg’s uncertainty principle :

Δx×mΔv=h4π

Δx=h4πmΔV ΔV=0.005100×600=0.03

Δx=6.6×10344×3.14×9.1×1031×0.03

=1.92×103 m

Correctoption is (c)

Question 14. Ionisation energy of He+is 19.6×1018 Jat1. The energy of the stationary state (n=1) of Li2+ is

(a) 4.41×1016Jatom1

(b) 2.2×1015Jatom1

(c) 4.41×1017Jatom1

(d) 8.82×1017Jatom1

Show Answer

Answer (c)

IE=E1

(E1)Het (E1)i+2i+=(ZHet )2(ZLi+2)2

19.6×1018(E1)Li2+=(ZHet )2(ZLi2)2

19.6×1018(E1)Li2+=49

Or(E1)Li2+=19.6×1018×94

=4.41×1017Jatom1

Correct option is (c)

Question 15. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at

(a) 1035 nm

(b) 743 nm

(c) 325 nm

(d) 518 nm

Show Answer

Answer (b)

hcλ=hcλ1+hcλ2

Or 1λ=1λ1+1λ2

1355=1680+1λ2

Or 1λ2=13551680

=680355680×355

=1.346×103

Or λ2=743 nm

PRACTICE QUESTIONS

1. Rutherford’s experiment on scattering of α-particles showed for the first time that the atom has

(a) electrons

(b) protons

(b) nucleus

(d) neutrons

Show Answer Answer: (c)

2. The increasing order [lowest first] for the values of e/m [charge / mass] for electron [e], proton [p], neutron [n] and alpha particles [α] is

(a) e,p,n,α

(b) n,p,e,α

(c) n,p,α,e

(d) n,α,p,e

Show Answer Answer: (d)

3. Bohr’s model can explain

(a) the spectrum of hydrogen atom only

(b) spectrum of atom or ion containing one electron only

(c) the spectrum of hydrogen molecule

(d) the solar spectrum

Show Answer Answer: (a)

4. Rutherford’s alpha particle scattering experiment eventually led to the conclusion that

(a) mass and energy are related

(b) electrons occupy space around the nucleus

(c) neutrons are buried deep in the nucleus

(d) the point of impact with matter can be precisely determined

Show Answer Answer: (b)

5. The wavelength of the electron emitted, when in a hydrogen atom, electron falls from infinity to stationary state 1 , would be [Rydberg’s constant =1.097×107 m1 ]

(a) 91 nm

(b) 192 nm

(c) 406 nm

(d) 9.1×108 nm

Show Answer Answer: (a)

6. The shortest wavelength in hydrogen spectrum of lyman series, when RH=109678 cm1 is

(a) 1002.7\AA

(b) 1215.67\AA

(c) 1127.30\AA

(d) 911.7\AA

Show Answer Answer: (d)

7. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to 8/9 times the Rydberg’s constant if the electron jumps from

(a) n=3 to n=1

(b) n=10 to n=1

(c) n=9 to n=1

(d) n=2 ton =1

Show Answer Answer: (a)

8. Those species are called isotones which have same

(a) atomic number

(b) mass number

(c) number of electrons

(d) number of neutrons

Show Answer Answer: (d)

9. Calculate the wavelength of the light required to break the bond between two chlorine atoms in a chlorine molecule. The ClCl bond energy is 243 kJ mol1

(a) 8.18×1031 m

(b) 6.26×1021 m

(c) 4.91×107 m

(d) 4.11×106 m

Show Answer Answer: (c)

10. A 600 W mercury lamp emits monochromatic radiation of wavelength 3.315×107 m. How many photons are emitted from the lamp per second?

(a) 1×109

(b) 1×1020

(c) 1×1021

(d) 1×1023

Show Answer Answer: (c)

11. Photoelectric emission is observed from a metal surface with incident frequencies v1 and v2 where v1>v2 If the kinetic energies of the photoelectrons emitted in the two cases are in the ration 2:1, then the threshold frequency v0 of the metal is

(a) v1v2

(b) v1v2h

(c) 2v1v2

(d) 2v2v1

Show Answer Answer: (d)

12. The ratio of the energy of a photon of 2000\AA\AA wavelength radiation to that of 4000\AA radiation is

(a) 1/4

(b) 4

(c) 1/2

(d) 2

Show Answer Answer: (d)

13. Atom of an element has Z electrons and its atomic mass is 2Z+3. The number of neutrons in its nucleus will be

(a) 2Z

(b) Z+3

(c) Z+2

(d) Z

Show Answer Answer: (b)

14. If an isotope of hydrogen has two neutrons in its atom, its atomic number and mass number will be

(a) 2 and 1

(b) 3 and 1

(c) 1 and 1

(d) 1 and 3

Show Answer Answer: (d)

15. The triad of nuclei that is isotonic is

(a) 614C,715 N,917 F

(b) 612C,714 N,919 F

(c) 614C,714 N,917 F

(d) 614C,714 N,919 F

Show Answer Answer: (a)

16. Which of the following is deflected most under the effect of electric field ?

(a) α-rays

(b) β-rays

(c) γ-rays

(d) X-rays

Show Answer Answer: (b)

17. The frequency of light having wavelength 500 nm is

(a) 5×1015 Hz

(b) 5×1010MHZ

(c) 2×1015 Hz

(d) 6×1014 Hz

Show Answer Answer: (d)

18. The ionization energy of He+is 19.6×1018 J/ atom. The energy of the first stationary state of Li+2 is

(a) 2.18×1018 J/ atom

(b) 4.90×1018 J/ atom

(c) +4.90×1018 J/ atom

(d) +4.41×1017 J/ atom

Show Answer Answer: (d)

19. The energy of an electron in the first orbit of hydrogen is 13.6eV. Which one of the following is the second excited state of electron in hydrogen atom?

(a) 3.4eV

(b) 6.8eV

(c) 1.5eV

(d) +3.4eV

Show Answer Answer: (c)

20. The wave number of the radiation emitted when the electron jumps from fourth energy level to second energy level in He+is about

(a) 20565 cm1

(b) 41030 cm1

(c) 82258 cm1

(d) 5141 cm1

Show Answer Answer: (a)

21. The energy of the second Bohr’s orbit of the hydrogen atom is 328 kJ mol1. The energy of fourth Bohr’s orbit would be

(a) 1312 kJ mol1

(b) 164 kJ mol1

(c) 82 kJ mol1

(d) 41 kJ mol1

Show Answer Answer: (c)

22. The threshold frequency of a metal is 1×1015 s1. The ratio of maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5×1015 s1 and 2.0x 1015 s1 respectively, would be

(a) 3:4

(b) 1:2

(c) 2:1

(d) 4:3

Show Answer Answer: (b)

23. The threshold frequency of a metal ’ M ’ is 1.5×1015 s1. The maximum K.E. of the photoelectrons, when the metal is irradiated with radiations of frequency 2.5×1015 s1, would be

(a) 6.63×1020 J

(b) 1.0×1019 J

(c) 6.63×1019 J

(d) 1.0×1018 J

Show Answer Answer: (c)

24. The de-Broglie wavelength associated with a body of mass 1000 g moving with a velocity 100 ms1 is

(a) 6.62×1039 m

(b) 6.62×1036 cm

(c) 6.62×1036 m

(d) 3.31×1032 m

Show Answer Answer: (c)

25. The H-specturm shows

(a) Heisenberg’s uncertainity principle

(b) diffraction

(c) polarisation

(d) presence of quantized energy levels

Show Answer Answer: (d)

26. Correct set of four quantum numbers for the valence (outermost) electron of rubidium ( Z=37) is

(a) 5,0,0,+12

(b) 5,1,0,+12

(c) 5,1,1,+12

(d) 6,0,0,+12

Show Answer Answer: (a)

27. The principal quantum number of an orbital is related to the

(a) size of the orbital

(b) spin angular momentum

(c) orientation of the orbital in space

(d) orbital angular momentum

Show Answer Answer: (a)

28. Which one of the following sets of quantum numbers represents an impossible arrangement?

n1ms(a)32212(b)40012(c)32312(d)53012

Show Answer Answer: (c)

29. Which of the following has the maximum number of unpaired electrons?

(a) Mg2+

(b) Ti3+

(c) V3+

(d) Fe2+

Show Answer Answer: (d)

30. The number of nodal planes in a px orbital is

(a) one

(b) two

(c) three

(d) zero

Show Answer Answer: (d)

31. The electrons, identified by quantum numbers n and I, (i) n=4,I=1, (ii) n=4,I=0, (iii) n=3,I=2, and (iv) n=3,I=I can be placed in order of increasing energy, from the lowest to highest, as

(a) (iv) < (ii) < (iii) < (i)

(b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (iv)

(d) (iii) < (i) < (iv) < (ii)

Show Answer Answer: (a)

32. The orbital angular momentum of an electron is 2 s orbital is

(a) +12h2π

(b) zero

(c) h2π

(d) 2h2π

Show Answer Answer: (b)

33. The number of radial nodes in 3s and 2p orbitals, respectively are

(a) 2 and 0

(b) 0 and 2

(c) 1 and 2

(d) 2 and 1

Show Answer Answer: (a)

34. The correct set of quantum numbers for the unpaired electron of chlorine atom is

n1mn1m(a)210(b)211(c)311(d)300

Show Answer Answer: (c)

35. Maximum number of electrons in a subshell of an atom is determined by the following

(a) 21+1

(b) 412

(c) 2n2

(d) 41+2

Show Answer Answer: (d)

36. What is the maximum number of electrons that can be associated with the following set of quantum numbers n=2,I=0, m=0

(a) 10

(b) 6

(c) 4

(d) 2

Show Answer Answer: (d)

37. The following quantum number are possible for how many orbitals? n=3,I=2,m=+2

(a) 1

(b) 2

(c) 3

(d) 4

Show Answer Answer: (a)

38. If n=3,I=0,m=0, then atomic number is

(a) 12,13

(b) 13,14

(c) 10,11

(d) 11,12

Show Answer Answer: (d)

39. The number of electrons is an atom with atomic number 105 having (n+I)=8 is

(a) 15

(b) 17

(c) 19

(d) 21

Show Answer Answer: (b)

40. Which of the following set of quantum numbers is not possible?

(a) n=3,I=2,m=0,s=1/2

(b) n=3,I=2,m=2,s=1/2

(c) n=3,I=3,m=3,s=+12

(d) n=3,I=1,m=0,s=+1/2

Show Answer Answer: (c)

41. According to Bohr’s theory, angular momentum for an electron is 5th orbit is:

(a) 5 hπ

(b) 2.5 hπ

(c) 5 h

(d) 25hπ

Show Answer Answer: (b)

42. A5f orbital has

(a) one node

(b) two nodes

(c) threenodes

(d) four nodes

Show Answer Answer: (d)

43. The azimuthal quantum number for 17 th electron of Cl atom is

(a) 1

(b) 2

(c) 3

(d) 0

Show Answer Answer: (a)

44. If electronic structure of oxygen atom is written as 1s12s2 , if would violate

(a) Hund’s rule

(b) Pauli’s exclusion principle

(c) both a and b

(d) none of the above

Show Answer Answer: (a)

45. With increase in principal quantum number ( n ), the energy difference between adjacent energy levels in hydrogen atom

(a) increases

(b) decreases

(c) remains contant

(d) decreases for lower values of n and increases for higher values of n

Show Answer Answer: (b)

46. The de-Broglie wavelength of an electron which has kinetic energy of 2.14×1022 J would be (me =9.1×1031 kg )

(a) 9.3×104 m

(b) 9.3×107 m

(c) 9.3×108 m

(d) 9.3×1010 m

Show Answer Answer: (c)

47. The nineteenth electron of chromium has which of the following set of quantum numbers?

(a) 3,0,0,1/2

(b) 3,2,2,1/2

(c) 4,0,0,1/2

(d) 4,1,1,12

Show Answer Answer: (c)

48. Which of the following statements is true for chloride ion and potassium ion, which are iso electronic?

(a) The have same size

(b) chloride ion is bigger than potassium ion

(c) Potassium ion is bigger than chloride ion

(d) Size of an ion depends on the other cation or anion present

Show Answer Answer: (b)

49. The configuration 1s2,2s2,2p53s1 represents

(a) Ground state of N3

(b) Ground state of F

(c) Excited state of O2

(d) Excited state of neon

Show Answer Answer: (d)

50. The ground state electronic configuration of chromium is

(a) [Ar]3 d54 s1

(b) [Ar]3 d44 s2

(c) [Ar]3 d64 s0

(d) [Ar]4 d54 s1

Show Answer Answer: (a)

51. What will be the uncertainty in the momentum of an electron if uncertainty in its position is zero

(a) Zero

(b) h4π

(c) <h4π

(d) infinite

Show Answer Answer: (d)

52. The electrons distribute themselves to retain similar spins as far as possible in a set of degenerate orbitals. This statement relates to

(a) Pauli’s Exclusion Principle

(b) Law of degeneration

(c) Hurd’s Rule

(d) Aufbau Principle

Show Answer Answer: (c)

53. The electron with which of the following values of quantum numbers is expected to possess highest energy

(a) 3,2,2,1/2

(b) 4,2,0,1/2

(c) 4,1,0,1/2

(d) 5,0,0,1/2

Show Answer Answer: (b)

54. An electron with +112 value for spin quantum number and -1 value for magnetic quantum number can not be present in

(a) s-orbital

(b) p-orbital

(c) d-orbital

(d) f-orbital

Show Answer Answer: (a)

55. The maximum number of electrons that can be accommodated in a subshell is given by

(a) 21+1

(b) 4I+2

(c) 2n2

(d) 2I+2

Show Answer Answer: (b)

56. The ratio of ionization energy of H and Be3+ is

(a) 1:1

(b) 1:3

(c) 1:9

(d) 1:16

Show Answer Answer: (d)

57. The electrons present in K shell of any atom will differ in value of which quantum number

(a) Value of n

(b) Value of I

(c) Value of m

(d) Value of ms

Show Answer Answer: (d)

58. Cathode rays are

(a) Stream of electrons

(b) Stream of -particles

(c) Coloured radiations

(d) Electromagnetic radiations

Show Answer Answer: (a)

59. The de-Broglie wavelength for particles having same kinetic energy is

(a) Independent of its velocity

(b) Directly proportional to its velocity

(c) Inversely proportional to its velocity

(d) Unpredictable

Show Answer Answer: (b)

60. The electron level which allows hydrogen to absorb photons but not to emit is

(a) 1 s

(b) 2 s

(c) 2p

(d) 3 d

Show Answer Answer: (a)

61. Which of the following ions has the maximum magnetic moment?

(a) Mn2+

(b) Ti2+

(c) Fe2+

(d) Cr2+

Show Answer Answer: (a)

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

(a) 25

(b) 23

(c) 52

(d) 32

Show Answer Answer: (c)

63. In a multi electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic field?

(i) n=1,I=0,m=0

(ii) n=2,I=1, m=1

(iii) n=2,I=0, m=0

(iv) n=3,I=2, m=0

(v) n=3,I=2,m=1

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (iii) and (iv)

(d) (iv) and (v)

Show Answer Answer: (d)

64. The uncertainties in the velocity of two particles A and B are 0.05 and 0.02 ms1 respectively. The mass of B is five times to that of A. The ratio of uncertainties in their positions is :

(a) 2

(b) 0.25

(c) 4

(d) 1

Show Answer Answer: (a)

65. What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? ( h= Planck’s constant, c= velocity of light, R= Rydberg’s constant)

(a) 5hcR36

(b) 4hcR3

(c) 3hcR4

(d) 7hcR144

Show Answer Answer: (c)

66. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103 ms1. (mass of proton =1.67×1027 kg and h=6.63×1034Js )

(a) 0.032 nm

(b) 0.40 nm

(c) 2.5 nm

(d) 14.0 nm

Show Answer Answer: (b)

67. The energy required to break one mole of ClCl bonds in Cl2 is 242 kJ mol1. The longest wavelength of light capable of breaking a single ClCl bond is

(a) 594 nm

(b) 640 nm

(c) 700 nm

(d) 494 nm

Show Answer Answer: (d)

68. The frequency of light emitted for the transition n=4 to n=2 of He+is equal to the transition in H atom corresponding to which of the following?

(a) n=2 to n=1

(b) n=4 to n=3

(c) n=3 to n=2

(d) n=3 to n=1

Show Answer Answer: (a)

69. Which of the following statements is correct for an electron that has the quantum numbers n=4 and m=2 ?

(a) the electron may be in d orbital.

(b) the electron may be in p orbital

(c) the electron is in the second principal shell.

(d) the electron must have a spin +12.

Show Answer Answer: (a)

70. Energy of an electron is given by

E=2.178×1018 J(Z2n2)

Wavelength of light required to excite an electron in an Hydrogen atom from level n=1 to n=2 will be

(h=6.62×1034Js and c=3×108 ms1)

(a) 1.214×107 m

(b) 2.186×107 m

(c) 6.500×107 m

(d) 8.500×107 m

Show Answer Answer: (a)