DPP - Daily Practice Problems

Name : Date : I I
Start Time : End Time : I

CHEMISTRY (0
SYLLABUS: Atomic structure (Fundamental Particles, Atomic Models, Bohr'sAtomic Model ,
1
3]
Max. Marks : 120
Hydrogen spectrum, Sommerfield's Model) Time : 60 min.
GENERAL INSTRUCTIONS
• The Daily Practice Problem Sheet contains 30 MCQ's. For each question only one option is correct. Darken the correct drcle/
bubble in the Response Grid provided on each page.
• You have to evaluate your Response Grids yourself with the help of solution booklet.
• Each correct answer will get you 4 marks and 1 mark shall be deduced for each incorrect answer. No mark will be given/ deducted
if no bubble is filled. Keep a timer in front of you and stop immediately at the end of 60 min.
• The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that syllabus.
Refer syllabus sheet in the starting of the book for the syllabus of all the DPP sheets.
• pl l
After com eting the sheet check your answers with the solution booklet and comp ete the Result Grid. Fina y s pen l d time to
analyse your performance and revise the areas which emerge out as weak in your evaluation.

DIRECTIONS (Q.l-Q.21) : 'Fhcre are 21 multiple choice Q.3 The wavelengths of two ph otons are 2000A and 4000A
r espective y.
lWhat is the ratio of their en ergies?
questions. Each question has 4 choices (a), (b), (c) and (d),
out of which ONLY ONE choice is correct. (a) 114 (b) 4
Q.l For cathode rays, the value of elm - (c) l/2 (d) 2
(a) is independent of the nature of the cathode and the Q.4 Which type of radiation is not emitted by th e electronic
gas filled in the discharge tube structure of atoms?
(b) is co nstant (a) Ultraviolet light (b) X-rays
(c) is -1.7588 xto8 coulombs/g
(d) All of the above are correct (c) Visible light (d) y-Rays

Q.2 Arrange the following particles in increasing order of Q.5 An oil drop has 6.39 x I o-19 C charge. Find out the number
values of elm ratio : electron (e), proton (p), rneutron (n) and of electrons in this drop -
a.-particle (a.) - (a) 4 (b) 3
(a) n, p, e, a. (b) n, a., p, e
(c) n,p, o., e (d) e, p n , o.
, (c) 6 (d) 8

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Q.6 Find out the nwnber ofwavemade by a Bohr electron in one Q.14 The energy of an electron in the second and third Bohr
complete revolution in its Ydorbit ofhydrogen atom - orbits of the hydrogen atom is - 5.42 x 1 0- 12 ergs and
(a) 4 (b) 3 (c) 6 (d) 8 -2.41 x I o- 1 2 erg respectively. Calculate the wavelengt h of
+ 18
Q.7 The ionization energy ofHe is 1 9. 6 x 10- J atom- 1 . The the emitted radiation when the electron drops from third to
energy of the first stationary state ofLi2+ will be- second orbit -
(a) 2 J.2xJ0-18 J/atom (b) 44.1 0 x J0-18 J/atom (a) 5.6 x 103 A (b) 6.6 X 1 02· A
(C) 63 .2 X w-IS J/atOffi (d) 84.2 X w-IS J/atOill
3A 3
(c) 6.6 x 10 (d) l0.6 x l0 A
Q.8 The ionization energyofhydrogen atom is 13.6 eV. What Q.15 Find the number ofquanta of radiations of frequency
will be the ionization energy of He+ - 4.75 x 1 013 sec-1, required to melt 100 g of ice. The energy
(a) 13.6 eV (b) 27.2 eV (c) 54.4 eV (d) 122.4 eV required t·o melt 1 g of ice is 350 J -
Q.9 The ionization energy ofH-atom is 13.6 eV. The ionization (a) ) 1 13 X 1 Q2 0 (b) l 1}3X 1Q 18
energy ofLi+2 ion will be - 5
(c) l l l3 x l01 (d) llJ3xlQ21
(a) 13.6 eV (b) 27.2 eV (c) 54.4 eV (d) 122.4 eV
iS 4.4 X w- 1 9 J and bond energy per lllO)eCUle iS 4.0 X lQ-I9
Q.t6 Tbe energy absorbed by eacb molecule (A ) ofa substance
Q.IO Which transition of the hydrogen spectrum would have the 2
same length as the Bahner transition, n = 4 to n = 2 ofHe+
J. The kinetic energy ofthe molecule per atom will be:
(a) 2.2 x w- 1 9 J (b) 2 .0 x w- 19 1
spectrum?

(C) 4.0 X lQ-20 J

(a) � = 2ton 1 = 1 (b) n2 = 3 to n1 = l
(d) n2 = 5 to n 1 = 3 (d) 2.0 X J0-20 J
(c) n2 = 4 to n 1 = 2
Q.ll Given R = 1.0974 X 107 m- and h = 6.626 X I o-34 Js. The
1 Q.17 If an electron is present in n = 6 level. How many spectral
ionization energy of one mole of Li+2 ions will be as lines would be observed in case ofH atom?
follows- (a) lO (b) 15
(a) 11240 kJ mote-l (b) 11180 kJ mote-l (c) 20 (d) 25
(c) 12350 kJ mole- 1 (d) 1 5240 kJ mole-1 Q.18 Naturally occuring boron consists of two isotops whose
Q.12 Calculate the energy emitted when electrons of 1.0 g atom atomic weights are 10.01 and 1 1 .01. The atomic weight of
of hydrogen undergo transition giving the spectral line of natural boron is I 0.81. Calculate the percentage of each
lowest energy in the visible region of its atomic spectrum isotope in natural boron-

(RH = 1 . 1 X 107 m-1, = 3 X 108 ms- 1 , h = 6.62 X 1 Q-34

(a) 20,80 (b) 30,70
C
(c) 10,90 (d) 15,85
Js).
Q.19 From the following list of atoms, choose the no. ofpairs of
(a) 182.5 kJ (b) 132.5 kJ(c) 112.5 kJ(d) 122.5 kJ
isotopes, isobars and isotones respectively
Q.l3 The shortest wavelength in H spectrum of Lyman series
nU•t9K, 7 N , 80 6c •20ca , nU
1 6 0 3 9 K , 23 5 40 23 8
when R H = 109678 cm- 1 is - s
14 18 14 4o
•19 •

(a) 1215.67 A (b) 911.7 A (a) 3, 2, 2 (b) 2, 3, 2

(c) 1002.1 A (d) ll27.30 A (c) 2,2,3 (d) 2,2,2

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Q.20 Atomic raruus is ofthe order of I o- 8 em. and nuclear radius Q.24 Choose the correct options for hydrogen atom -
3
is ofthe order of Io-1 em. Calculate what fraction ofatom
(1) E2 = -
13 6 13 6
is occupied by nucleus? · eV (2) E6 =- . eV
(b) 10- 15 (c) 10- 1 (b) 10- 9
2 9 36
(a) 10- 10
Q.21 Nitrogen atom has atomic number 7 & oxygen has atomic 13 6
(3) E6 = - · eV
number 8. Calculate the total nwnberofelectrons in nitraet ion- 25
(a) 40 (b) 64 (c) 16 (d) 32 DIRECTIONS (Q.25-Q.27): Read the passage given below and
DIRECTIONS (Q.22-Q.24) : In the following questions, more answer the questions that follows :
than one of tbe answers given are correct. Select tbe correct SOMMERFIELD'S CONCEPT
answers and mark it accordingto the followingcooes: (a) Sommerfield in 1915, introdluced a newatomic modelto explain
Cooes : line spectrum of hydrogen atom
(a) 1 , 2 and 3 are correct (b) 1 and 2 are correct (b) He proposed that the moving electron might describe
(c) 2 and 4 are correct (d) 1 and 3 are correct elliptical orbits in addition to circular orbits, and the nucleus
is situated at one of the foci.
Q.22 For the table -
(c) During motion on a circle, only the angle of revolution
Atomic Mass changes while the distance from the nucleus remains the
Number (Z) No.
Atom/ion Protons Neutrons Electrons same but in elliptical motion both the angk of revolution

(e)
and the distance of the electron from the nucleus change.
(A) ( p) (n) (d) The distance from tbe nucleus is termed as radius vector
AIJ+ 13 X 14 and the angle of revolution is known as azimuthal angle.
(e) The tangential velocity oftlhe electron at a particular instant
Cu 29 63 y can be resolved into two components. One along the radius
Ma2+
0
12 24 z 12 vector called radial velocity and the other perpendicular to
the radius vector called transverse or angular velocity.
Choose the correct options -
(f) These two velocities give rise to radial momentum and
(I) x=27 (2) y=34 (3) z= 1 2 (4) z=25 <'Lilgular or azimuthal momentum.
Q.23 Choose the correct statements - (g) Sommerfield proposed that both the momenta must be
(I) The difference in energy between I st and 2nd Bohr orbit h
fora Hatom is+ 1 0.2 eV integral multiples, radial momentum = nr - , Azimuthal
21t
(2)At minimum atomic no. 2, a transition from n = 2 to
n = 1 energy level would result in the emission of h
X- ray with A = 3.0 x 1 0- 8 m.
momentum = nq,
21t
(3) The difference in energy between 1 st and 2nd Bohr orbit Q.25 To give designation to an orbital, we need -
for a Hatom is+ 12.1 eV (a) Principal and azimuthal quantum munbers
(4) At minimum atomic no. 4, a transition from n = 2 to (b) Principal and magnetic quantum numbers
n = 1 energy level would result in the emission of (c) Azimuthal and magnetic quantum numbers
X- ray with A = 3.0 x 1 0- 8 m. (d) Principal, azimuthal and magnetic quantum numbers

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Q.26 The elliptical orbits of electron in the atom were proposed (a) Statement-! is True, Statement-2 is True; Statement-2 is a
by - correct explanation for Statement-I .
(a) Thomson (b) Bohr (b) Statement-! is True, Statement-2 is True; Stat,ement-2 is NOT
(c) Sonunerfield (d) De Broglie a correct explanation for Statement-I .
Q.27 Choose the correct statements- (c) Statement - 1 is False, Statement-2 is True.
(a) Sommerfield model gives introduction of elliptical (d) Statement - 1 is True, Statement-2 is False.
orbitals. Q.28 Statement 1 : The atoms of different elements having same
(b) Energies ofsubshells follow the order s < p < d < f. mass number but different atomic number are known as
(c) The relation between principal (n) and azimuthal (e)
isobars.
Statement 2 : The sum of protons and neutrons, in the
m length of major axis isobars is always different.
quantum numbers is =
e length of rrtinor axis Q.29 Statement 1 : The value ofn for a line in Balmer series of
(d) All ofthese hydrogen spectrum having the highest wavelength is
4 and 6.
DIRECTIONS (Q.28-Q.30) : Each of these questions contains
two statements: Statement-! (Assertion) and Statement-2 Statement 2 : For Balmer series n = 2 and n2 = 3, 4, 5
1
(Reason). Each ofthese questions has four alternative choices, Q.30 Statement 1 : The transition ofelectrons n3 � n2 in H atom
only one of which is the correct answer. You have to select the will emit greater energy than n4 � n3.
correct choice. Statement 2 : n 3 and n2 are closer to nucleus than n4.

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DAILY PRACTICE PROBLEM SHEET 3 - CHEMISTRY

Tota l Questions 30 Tota l Marks 120
Attempted Correct
Incorrect Net Score
Cut-off Score 32 Qual ifying Score 52
Success Gap = Net Score - Qual ifying Score
N et Score = (Correct x 4} - (I ncorrect x 1 }

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DAILY PRACTICE C H E M I STRY

PROBLEMS SOLUTIONS
(I) (d) Cathode rays consist of electrons which are (8) (c) He+ is a hydrogen like species i.e., the electron is
fundamental particles ofmatter. ionised from first orbit.
z2 E
(2) (b) Electron Proton Neutron a -particle Ionization energy ofHe+= ----fi
e 1 unit 1 unit zero 2 units n
m I/1837 unit I unit I unit 4-units
4 X 13.6
elm 1837 1 zero l /2 2 =54.4eV
1
(9) (d) E, for u+ 2 = E, for H X Z2 [for Li, z = 3]
(3) (d) = 13.6 x 9 = 122.4eV
(a)

[�-�]
(10) For He+ ion, we have

_!_ = RH Z2
A n, n2

[-·1 ]
(4) (d) y-Rays emission occurs due to radioactive change,
J
a nuclear phenomenon. = RH [2)2 = �R H . . .(A)
(5)
_

Charge on an oil drop = 6.39 x 10- 1 9 C 22 42

_

(a) 4
Now we know that Now for H atom
1 .602 x w - 1 9 C is the charge on one 1 electron
. . 6.39 X IQ- 1 9 C charge will be On . . .(B)

6.39 X 10-!9 Equating eqs. (A) and (B), we have

- = 4 electrons
1 .602 X 10- !9 3
(6) (b) Weknowthat rn = ro X n2 11t 2 1122 4
,', [3 = 0.529 X w-8 Cill X (3? Obviously n 1 = 1 and n 2 = 2. Hence the transition
( ·: r0 = 0.529 x J0-8 em) n2 = 2 ton 1 = I in hydrogen atom will have the same
length as the transition n = 4 ton= 2 in He+ species.
Also we know that
(11) (b) The expression ofionization energy is :
uo
un = ­ .
: u3 =
2.19 x 108 �E = RZ2 hc
2
For Li+ ion, Z = 3, hence
�E= (1.0974 X 107 m- 1) X (9) X (6.626 X w-34 J.S.)
n 3
(·: u0 =2.19 x 1 08 em sec-1) x (3 x t08ms-1 ) = 1.964x w-1 7 J
No. ofwaves in one round For one mole ofions, we have
�E'= NA �E=(6.023 X 1023 mol- 1 )(1.964 X w- 17 J)
= 1 . 1 18 x 107 J mol- 1 = 1 1 180 kJ mol- 1
0

(12) (a) Tlhe spectral line lies in the visible region i.e., it
corresponds to theBalmer series i.e. n2 = 2 and hence
n 1 =3, 4, 5, etc.
Substituting the values of the different constants
No. ofwaves in one round For lowest energyofBalmer series, n 1 = 3

[ l
2 x 3. 1 4 x 0.529 x l0-8 x 9 Substituting the values in the following relation.

[1 ]
2
x 2.19 x l 08 x9. 108 x 1 0- 8
3 1
= RH 2 2 = 1 . 1 X 10 7 X ---
3x 6.62 x l 0-27
l I 1
J...
4 9
- - --

(7) (b) E, for Li+2 = E for H X Z2 = E for H X 9 02 nl

I Li I
E I for He+ = E I for H X z2 He = E I for H X 4 7
2 9 = l . l x i0 x _2_
or E I for Li+ = - E I for He+ 36
4
- = 6.55 x w - 7 m
9 36
= J 9.6 X w-! 8 X 4 = 44 1 0 X J 0-- !8 }/atOm - - -
l . l x l07 x 5
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c n(n - I)
Now, we know that, E = hv = h x "i (17) (b) The no. of spectral lines is given by
2
when n = 6, then the no. of spectral lines
6.62 X J0-34 X X 8 3 10 6 x (6 - l) = 6 x 5 = 15
6.55 X J0-7 =
2 2
:. Energy corresponding to lg atom of hydrogen
= 3 . 3 X J0- !9 X 6. 2 X 23
0 0 10 (18) (a) Letthepercentage ofisotopewithatomicwt. 10.01 = x
X
= 4 J = 182.5kJ
18.25 10 Percentage ofisotope with atomic wt. = 11 .0 1 100- x
(13) (b) For Lyman series, n 1 = I m1x 1 + m2x2
Average atomic wt. =
For shortest wavelength ofLyman series the energy
differnece in two levels showing transition should
or Average atomic wt.
be maximwn, (i.e., n =oo).
2

[__!_12 _ ---1 ] = 109678

x x 10.01 + (100 - x ) x l l.OI
A.
_!_ = RH 1 00
0<,2
X X 10.01+(100-x)xl l.Ol
:. A.= 9 t t.7 x w-8 = 9 t t .7 A 10.81 = �X = 20
1 27 IOO
(14) (c) Here, h = 6.62 x o- erg
% ·ofisotope with atomicwt. 10.01 =20
£3 = 2. 4 X IQ- 1 erg
- 1 2 %·ofisotope with at·omic wt. 11.01 = 100-x = 80
E2 = .42 X w- 12 .erg
-5 (19) (a) lsotopes :
- 2 - 41
t.E = E3 E = 2. x w- 12 + . 2 x 0- 1 54 1 2 ( 1 � 0 , 1 �0 ), ( �� K , i8K ), e�� U , 2g� u )
Now, we know that, �E = hv
4
2 Isobars : ( 4o K ' o c a ) ' c 147 N ' 146 c )
20
19
v - - - ---�
-
LlE
h
- 3.01 10-1
X

6.62 X J0-27 Isotones : ( 1�K , �gca ), ct�c , 1 � 0 )

Volumeofnucleus= (4/3)nr3
�
(20) (b)
. c c
SIUCe V = - '· rv = - = (4/3)n x ( I 0- 1 3? cm3
A v 3 cm3
Volume ofatom = 4/3 nr3 = (4/3)
nx( 1 o- 8)

6.62 X 10-27 X 3 X J08 3

vnuc1eus 1 0- 9
:. 'A = = = w- 15
3.0 1 x l0-12 Yatom 10-24
10-5cm
A. = 6.6 x or vnucleus = t o- I S X v tom
a
Since, lA= to-Bern (21) (d) No. of electrons in N03-
A.= 6.6 x 103 A = (Electrons in N) + (3 x electrons in 0)
(d) E = nhv = n x 6.62 x J0- 34 J sec x 4.75 x 1 01 3 sec-1 [1
+ (due to negative charge)]

= n x 31.445 x l o- 2 1 J =7 + 3 X
(15)
8+
I = 32
(22) (a)
Energyrequired tomelt 100gice=350Jx 100 (1) 13
Atomic number (Z) ofAI= = Number of protons
=350001
n X 3 .445 X I0-21 = 350QQ
1 Number ofelectrons= = 13-3 10
Mass number =n p = + 14+ 13 =27
n=
35000 = 11 13 x i02 1 (2) Atomic number = Number ofprotons
3 1 .445 x l 0-2 1 =Number ofelectrons = 29
Mass number = n + p = 63
(16) (d) K.E peratom
Since p = 29
j4.4 xi0-19)-{4.0 x ] 0- 19 ) :. 11 = 6 3 - p = 6 3 - 29 = 34
(3) Number ofprotons = Z = 1 2
2
Number ofelectrons= 1 2 - 2 = 10
0.4x 1 0- 19 Mass number = n + p = 24
= 2.0x l 0- 20 n = 24- p = 24- 1 2 = 1 2
2
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(23) (b) E 1 for H = - 13.6eV
E2 for H = ( - 1 3.6/22) = - 1 3.6/4 =- 3.4 eV
13.6 13.6 1 ( •) 2
E6 - E2 = 4- 6 = 13.6 4 - 3 6 = 13.6 X 9
3
E2 - E 1 =- 3 .4 - ( - 1 3.6) =+ 10.2 eV
Also for transition ofH like atom ; A. = 3.0 x 1 0- 8 m E2 - E1 > E6 - E2
(25) (d) The correct answer is (d).
(26) (c) The elliptical orbits of electron in the atom were
proposed by Sommerfield.
(27) (d) All statements are correct.
1 8 = 1 .09 x l07 x Z2 x l
(28) (d) Is·obars are the atoms ofdifferent elements having
4 same mass number but different atomic number, S-1
3 x w-
is correct but S-2 is false because atomic mass is sum
2
Z = 4 and Z = 2 ofnumber of neutrons and protons which should be
(24) (c) Energy ofn = I for H-atom same for isobars.
E 1 =-13.6eV (29) (c) We know that the line in Balmer series ofhydrogen
spectrum the highest wavelength or lowest energy
Energy ofn = 2 for H-atom
is between n1 = 2 and n2 = 3. And for Balmer series of
13.6
E2 = - 4 eV hydrogen spectrum, thevalueofn 1 = 2 and n2 = 3, 4,
5. Therefore t11e S-1 is false but the S-2 is true.
Energy of n = 6 for H-atom (30) Both statements are true, but S-2 is not the correct
13.6
(b)
E6 = - }6 eV explanation of S- 1 . The difference between the
energies of adjacent energy levels decreases as we
13.6 3 move always far from the nucleus. Thus in H atom
So, E2 - E 1 = 13.6 - 4 = 13.6 x 4 E2 - E 1 > E3 - E2 > E4 - E3 ..... .