Wednesday, 16 October, 2019

SOLVE QUESTIONS OF ATOMIC STRUCTURE – SOLVE PAPER – IIT-JEE/NEET/AIIMS/MCAT EXAM

Solved Examples on Atomic Structure - Study Material for IIT JEE ,ATOMIC STRUCTURE - SOLVE PAPER - NEET/AIIMS/MCAT EXAM

SOLVE QUESTIONS OF ATOMIC STRUCTURE – SOLVE PAPER – IIT-JEE/NEET/AIIMS/MCAT EXAM




Ex.1 Complete the following –
Atom/ion Atomic Mass Proton Neutrons Electrons
Number (Z) No.
(A) (p) (n) (e)
Al3+ 13 14
Cu 29 63
Mg2+ 24 12
Sr 88 38
Sol. (i) Atomic number (Z) = 13 = Number of protons
Number of electrons = 13 –3 = 10
Mass number = n + p = 14 + 13 = 27
(ii) Atomic number = Number of protons
= Number of electrons = 29
Mass number = n + p = 63
since p = 29
\ n = 63 – p = 63 – 29 = 34
(iii) Number of protons = Z = 12 & Number of electrons = 12 – 2 = 10
Mass number = n + p = 24
\ n = 24 – p = 24 – 12 = 12
(iv) Number of electrons = Number of protons = Z = 38
Mass number = n + p = 88
\ n = 88 – p = 88 – 38 = 50

Ex.2 An oil drop has 6.39 × 10– 19 C charge. Find out the number of electrons in this drop –
Sol. Charge on oil drop = 6.39 × 10– 19 C
Now we know that
1.602 × 10– 19 C is the charge on one 1 electron
\ 6.39 × 10– 19 C will be charge on = = 4 electrons

Ex.3 The ionization energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary state of Li+2 will be –
(A) 21.2 × 10–18 J/atom (B) 44.10 × 10–18 J/atom
(C) 63.2 × 10–18 J/atom (D) 84.2 × 10–18 J/atom (Ans. B)
Sol. E1 for Li+2 = E1 for H × Z2 Li = E1 for H × 9
E1 for He+ = E1 for H × Z2 He = E1 for H × 4
or E1 for Li+2 = E1 for He+
= 19.6 × 10–18 ×
= 44.10 × 10–18 J/atom
Ex.4 The ionization energy of hydrogen atom is 13.6 eV. What will be the ionization energy of He+ –
Sol. He+ is a hydrogen like species i.e. the electron is ionised from first orbit.
\ Ionization energy of He+ =
= = 54.4 eV
Ex.5 Which transition of the Hydrogen spectrum would have the same lenght as the Balmer transition, n = 4 to n = 2 of He+ spectrum –
(A) n2 = 2 to n1 = 1 (B) n2 = 3 to n1 = 1 (C) n2 = 4 to n1 = 2 (D) n2 = 5 to n1 = 3 (Ans. A)
Sol. For He+ ion, we have
=
=
= …(A)
Now for H atom
= …(B)
Equating equs (A) and (B) we have
=
Obviously n1 = 1 and n2 = 2. Hence the transition n = 2 to n = 1 in hydrogen atom will have the same length as the transition n = 4 to n = 2 in He+ species.

Ex.6 The shortest wave length in H spectrum of Lyman series when RH = 109678 cm–1 is –
(A) 1215.67 Å (B) 911.7 Å (C) 1002.7 Å (D) 1127.30 Å (Ans B)
Sol. For Lyman series n1 = 1
For shortest ‘l’ of Lyman sereis the energy differnece in two levels showing transition should be maximum (i.e. n2 = ¥).
= RH
= 109678
\ l = 911.7 × 10– 8
= 911.7 Å

Ex.7 Find the number of quanta of radiations of frequency 4.75 × 1013 sec–1, required to melt 100 g of ice. The energy required to melt 1 g of ice is 350 J –
Sol. E = nhv
= n × 6.62 × 10– 34 J sec × 4.75 × 1013 sec–1
= n × 31.445 × 10– 21 J
Energy required to melt 100 g ice = 350 J × 100
= 35000 J
n × 31.445 × 10– 21 = 35000
n = = 1113 × 1021



 

Ex.8 Calculate the wavelength of a moving electron having 4.55 × 10– 25 J of kinetic energy –
Sol. Kinetic energy = (½mu2) = 4.55 × 10–23 J
\ u2 =
\ u = 103 m sec–1
\ l =
= 7.27 × 10– 7 meter

Ex.9 Show that the wavelength of a 150 g rubber ball moving with a velocity 50 m sec–1 is short enough to be observed –
Sol.
Given u = 50 m sec–1
= 50 × 102 cm sec– ; m = 150 g
\ l = = 8.83 × 10– 33 cm
The wavelength is much longer than the l of visible region and thus it will not be visible.
Ex.10 If an electron is present in n = 6 level. How many spectral lines would be observed in case of H atom–
(A) 10 (B) 15 (C) 20 (D) 25 (Ans B)
Sol. The no. of spectral lines is given by
when n = 6 then, the no. of spectral lines
=
Ex.11 What designation will you assign to an orbital having following quantum number –
(a) n = 3, l = 1, m = –1
(b) n = 4, l = 2, m = +2
(c) n = 5, l = 0, m = 0
(d) n = 2, l = 1, m = 0

Sol. (a) Since l = 1 corresponds to p-orbital and m = –1 shows orientation either in x or y axis, thus this
orbital refers to 3px or 3py
(b) or
(c ) 5s
(d) 2pz
Ex.12 How many electrons in a given atom can have the following quantum numbers –
(a) n = 4, l = 1 (b) n = 2, l = 1, m = – 1, s = + ½
(c) n = 3 (d) n = 4, l = 2, m = 0
Sol. (a) l = 1 refers to p – subshell which has three orbitals (px, py and pz) each having two electrons. Therefore, total number of electrons are 6.
(b) l = 1 refers to p – subshell, m = – 1 refers to px or py orbital whereas, s = +½ indicate for only 1 electron.
(c) For n = 3, l = 0, 1,
2 l = 0 m = 0 2 electrons
l = 1 m = –1 6 electrons
l = 2 m = –2 , –1, 0, +1, +2 10 electrons
Total electrons 18 electrons
Alternatively, number of electrons for any energy level is given by
2n2 i.e. 2 × 32 = 18 electrons
(d) l = 2 means d-subshell and m = 0 refer to dz2 orbital
\ Number of electrons are 2.
Ex.13 Which of the following set of quantum numbers are not permitted –
(a) n = 3, l = 2, m = – 1, s = 0
(b) n = 2, l = 2, m = +1, s = – ½
(c) n = 2, l = 2, m = + 1, s = – ½
(d) n = 3, l = 2. m = – 2, s = + ½
Sol. (a) This set of quantum number is not permitted as value of ‘s’ cannot be zero.
(b) This set of quantum number is not permitted as the value of ‘l ’ cannot be equal to ‘n’.
(c) This set of quantum number is not permitted as the value of ‘l ’ cannot be equal to ‘n’.
(d) This set of quantum number is permitted.
Ex.14 Naturally occuring boron consists of two isotops whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. Calculate the percentage of each isotope in natural boron-
Sol. Let the percentage of isotope with atomic wt. 10.01 = x
\ Percentage of isotope with atomic wt.
11.01 = 100 – x
Average atomic wt. =
or Average atomic wt. =
10.81 = x = 20
\ % of isotope with atomic wt. 10.01 = 20
% of isotope with atomic wt. 11.01 = 100 – x = 80

Ex.15 From the following list of atoms, choose the isotopes, isobars and isotones –
, , , , , , , ,
Sol. Isotopes :
(, ), (,), (,)
Isobars : (, ) , (, )
Isotones : (, ), (, )

Ex.16 Atomic radius is the order of 10– 8 cm. and nuclear radius is the order of 10–13 cm. Calculate what fraction of atom is occupied by nucleus –
Sol. Volume of nucleus = (4/3)pr3
= (4/3)p × (10–13)3 cm3
volume of atom = 4/3 pr3 = (4/3) p × (10– 8)3 cm3
\
or Vnucleus = 10– 15 × Vatom
Ex.17 If the mass of neutrons is doubled & mass of electrons is halved then find out the atomic mass of 6Carbon12 and the percentage by which it is increased.
Sol. Step-1 6C12
e = 6
P = 6 = 6 amu
n = 6 = 6 amu = 12 amu
It the mass of neutrons is doubled and mass of e– is halved then.
n = 12 amu
p = 6 amu = 18 amu
Note : mass of e– is negligible, so it is not considered in atomic mass.
Step-2 % by which atomic mass is increased is x, then
12 × = 6 [18 – 12 = 6 amu]
x =
x =
x = 50 %

 



Ex.18 Calculate the frequency & energy of a photon of wave length 4000 Aº
Sol. (a) Calculation of frequency
l = 4000 Aº
l = 4000 × 10–10m
Q v =
\ v =
= 0.75 ×1015–sec–1
= 7.5 × 1014 sec–1
(b) Calculation of energy :
E = hv
= 6.266 ×10–34 Joule × 7.5 ×1014 sec–1
= 4.96 ×10–19 Joule
Ex.19 If the P.E. of an electron is –6.8 ev in hydrogen atom then find out K.E., E orbit where electrons exist & radius of orbit.
Sol. 1. P.E. = –2 K.E.
– 6.8 = –2 K.E.
= K.E.
K.E. = 3.4 ev
2. E. = – K.E.
= –3.4 ev
3. Orbit = 2nd
\ E = –13.6 ×
\ 3.4 = –13.6 ×
Þ n2 = = 4
i.e. n = 2
4. r = 0.529 × Aº
r = 0.529 ×
= 0.529 × 4Aº = 2.116 Aº

Ex.20 Calculate the wavelength of 1st line of Balmer series in Hydrogen spectrum.
Sol. For first line of Balmer series
n1 = 2, n2 = 3
= R(1)2
= R
= R
l =
l =
= × 9.12 ×10–6 cm
= 65.66 ×10–6 Cm
{\ 1Aº = 10–8cm, 1Cm = 108 Aº }
= 65.66 × 10–6 ×10–8 Aº
= 65.66 × 102Aº
= 6566 Aº

Ex.21 Calculate the frequency of the line of lyman series in hydrogen spectrum. For last line of lyman series n1 = 1, n2 = ¥
Sol. =
= R
= R
= 109678Cm–1
v = = C × = C × R
= 3 × 1010 × 109670
= 3.29× 1015 sec–1

Ex.22 What should be the momentum (in gram cm per second) of particle Its De Broglie Wavelength is 1 Å and the value of h is 6.6252 ×10–27 erg second ?
(A) 6.6252 ×10–19 (B) 6.6252 ×10–21
(C) 6.6252 ×10–24 (D) 6.6252 ×10–27
Sol. Given that
l = 1 Å = 1 ×10–8cm
h = 6.6252 × 10–27 erg second
or p =
= 6.6252 ×10–19 gram cm/sec.



 

Ex.23 How many photons of lights having a wave length of 5000 Aº are necessary to procide 1 Joule of energy.
Sol. Q E =
\ n =
= × 3 × 108
= 2.5 × 1018 photons
Ex.24 Calculate the radius ration of 2nd excited state of H & 1st excited state of Li+2
Sol. 2nd excited state, means e– is present in 3rd shell of hydrogen
r3 = 0.529 ×
= 0.529 × 9
1st excited state, means e– exist in 2nd shell of Li+2
r2 = 0.529 ×
= 0.529 ×
=
=
= =
Ex.25 A certain electronic transition from and excited state to Ground state of the Hydrogen atom in one or more steps gives rise of 5 lines in the ultra violet region of the spectrum. How many lines does this transition produce in the Infra red region of the spectrum ?
Sol. (Lyman Series) ultral violet region :
5 Lines i.e. e– is coming from 6th to 1st Orbit
n2 – 1 = 5
n2 = 6
Infrated region line
(i) Paschen series = (6 – 3) = 3
(ii) Bracket = (6 – 4) = 2
(iii) Pfund = (6 – 5) = 1
Total Number of Lines are = 6
Ex.26 What should be the ratio of velocities of CH4 and O2 molecules so that they are associated with de Brogile waves of equal wave lengths ?
Sol. According to de Broglie equation = l =
for methane (CH4) =
For oxygen (O2) =
Since the wavelenght of CH4 and O2 are to be equal
=
or =

or = = = 2

Ex.27 Calculate the wavelength, frequency and wave number of a light wave period is 2.0 × 10–10 s.
Sol. Frequency = =
= 5.0 × 109 s–1
Wavelength = = = 6.0 × 10–2m
Wave no. = = = 16.66 m–1
Ex.28 How much energy is required to ionise a H-atom if the electron occupies n = 5 orbit ? Compare your answer with the ionization energy of H-atom (energy eqquired to remove the electron from
n = 1 orbit).
Sol. E1 = –13.6 eV ;
E5 = = 0.544eV
= 0.544 × 1.6022 ×10–19J = 8.71 ×10–20J
thus, 0.544 eV energy is required to ionised H atom if electron is in 5th obit.
Also = = 0.04

Ex.29 (i) The energy associated with the first orbit in the hydrogen atom is –2.17 × 10–18 J atom–1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Sol. (i) E1 = – 2.17 × 10–18 J
E5 = = = –8.68 ×10–20 J
(ii) Rn = r1 × n2 = 0.529 × 52
= 13.225 Å
= 1.3223 nm.

 



Atomic structure Theory

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