Atoms Practice Questions - DPP for NEET

 Page 1


1. (a) For 
1 n =
, maximum number of states 
2
22 n == and
for 2,3,4 n = maximum number of states would be 8,
18, 32 respectively , Hence number of possible elements
2 8 18 32 60 = ++ +=
2. (d) 2
2
2
1
1 11
2
RZ
æö
= ç -÷
ç÷
l
èø
For di-ionised lithium the value of  Z is maximum
3. (c) Lyman series lies in the 
UV
region.
4. (c) Transition A (n =¥ to 1): Series limit of Lyman series
Transition B (5 n = to 2) n = Third spectral line of
Balmer series
Transition C (5 n = to 
3 n =
) : Second spectral line of
Paschen series
5. (b) Let the energy in A, B and C state be E
A
, E
B
 and E
C
then from the figure
1
l
2
l
3
l
( )( )()
C B B A CA
E E E E EE -+- = -
or
1 23
hc hc hc
+=
l ll
12
3
12
ll
Þl=
l +l
6. (a) In the revolution of electron, coulomb force provides
the necessary centripetal force
222
2
2
Ze mv Ze
mv
rr
r
Þ = Þ=
\
K.E. 
2
2
1
22
Ze
mv
r
==
+
e
r
7. (a) P .E. 
1
r
µ- and K.E. 
1
r
µ
As r increases so K.E. decreases but P .E. increases.
8. (a) K.E. =- (T.E.)
9. (a) For Lyman series
22
max
1 13
4
(1) (2)
Lyman
c RC
v Rc
éù
= = -=
êú
l
êú
ëû
For Balmer series
22
max
1 15
36
(2) (3)
Balmer
c RC
v Rc
éù
= = -=
êú
l
êú
ëû
27
5
Lyman
Balmer
v
v
\=
10 (c)
22
12
1 11
R
nn
æö
=- ç÷
ç÷
l
èø
For first line of Lymen series 
1
1 n = and 
2
2 n =
For first line of Balmer series 
2
2 n = and 
2
3 n =
So, 
5
27
Lymen
Balmer
l
=
l
11. ( d) Angular momentum 
2
h
Ln
æö
=
ç÷
p
èø
For this case 
2 n =
, hence 
2
2
hh
L=´=
pp
12. (c)
22
2
00 00 0
1
4 4
mv ee
v
a am a
= Þ=
pe pe
13. (d) We have to find the frequency of emitted photons. For
emission of photons the transition must take place from
a higher energy level to a lower energy level which are
given only in options (c) and (d).
Frequency is given by
22
21
11
13.6 h
nn
æö
n=--
ç÷
ç÷
èø
For transition from n = 6 to n = 2,
1
22
13.6 1 1 2 13.6
9
62
hh
æö - æö
n= - =´
ç÷
ç÷
èø
èø
For transition from n = 2 to n = 1,
2
22
13.6 1 1 3 13.6
4
21
hh
æö - æö
n = - =´
ç÷ ç÷
èø èø
.
\
 
21
n >n
Page 2


1. (a) For 
1 n =
, maximum number of states 
2
22 n == and
for 2,3,4 n = maximum number of states would be 8,
18, 32 respectively , Hence number of possible elements
2 8 18 32 60 = ++ +=
2. (d) 2
2
2
1
1 11
2
RZ
æö
= ç -÷
ç÷
l
èø
For di-ionised lithium the value of  Z is maximum
3. (c) Lyman series lies in the 
UV
region.
4. (c) Transition A (n =¥ to 1): Series limit of Lyman series
Transition B (5 n = to 2) n = Third spectral line of
Balmer series
Transition C (5 n = to 
3 n =
) : Second spectral line of
Paschen series
5. (b) Let the energy in A, B and C state be E
A
, E
B
 and E
C
then from the figure
1
l
2
l
3
l
( )( )()
C B B A CA
E E E E EE -+- = -
or
1 23
hc hc hc
+=
l ll
12
3
12
ll
Þl=
l +l
6. (a) In the revolution of electron, coulomb force provides
the necessary centripetal force
222
2
2
Ze mv Ze
mv
rr
r
Þ = Þ=
\
K.E. 
2
2
1
22
Ze
mv
r
==
+
e
r
7. (a) P .E. 
1
r
µ- and K.E. 
1
r
µ
As r increases so K.E. decreases but P .E. increases.
8. (a) K.E. =- (T.E.)
9. (a) For Lyman series
22
max
1 13
4
(1) (2)
Lyman
c RC
v Rc
éù
= = -=
êú
l
êú
ëû
For Balmer series
22
max
1 15
36
(2) (3)
Balmer
c RC
v Rc
éù
= = -=
êú
l
êú
ëû
27
5
Lyman
Balmer
v
v
\=
10 (c)
22
12
1 11
R
nn
æö
=- ç÷
ç÷
l
èø
For first line of Lymen series 
1
1 n = and 
2
2 n =
For first line of Balmer series 
2
2 n = and 
2
3 n =
So, 
5
27
Lymen
Balmer
l
=
l
11. ( d) Angular momentum 
2
h
Ln
æö
=
ç÷
p
èø
For this case 
2 n =
, hence 
2
2
hh
L=´=
pp
12. (c)
22
2
00 00 0
1
4 4
mv ee
v
a am a
= Þ=
pe pe
13. (d) We have to find the frequency of emitted photons. For
emission of photons the transition must take place from
a higher energy level to a lower energy level which are
given only in options (c) and (d).
Frequency is given by
22
21
11
13.6 h
nn
æö
n=--
ç÷
ç÷
èø
For transition from n = 6 to n = 2,
1
22
13.6 1 1 2 13.6
9
62
hh
æö - æö
n= - =´
ç÷
ç÷
èø
èø
For transition from n = 2 to n = 1,
2
22
13.6 1 1 3 13.6
4
21
hh
æö - æö
n = - =´
ç÷ ç÷
èø èø
.
\
 
21
n >n
DPP/ P 55
154
14. (d)
2
13.6 9 13.6 122.4 E Z eV eV eV =-´=-´=-
So ionisation energy 122.4eV =+
15. (c) For third line of Balmer series 
12
2,5 nn ==
2
22
12
1 11
RZ
nn
éù
\=- êú
l
êú
ëû
gives 
22
2 12
22
21
()
nn
Z
n nR
=
-l
On putting values Z = 2
From 
22
22
13.6 13.6(2)
54.4
( 1)
Z
E eV
n
-
=- = =-
16. (b)
2 2 22
32
12
1 1 1 1 115
36
(2) (3)
R
RR
nn ®
éù éù
= - Þ = -= êú êú
ll
êú êú
ëû ëû
and 
22
42
1 1 13
16
(2) (4)
R
R
®
éù
= -=
êú
l
êú
ëû
42
420
32
20 20
27 27
®
®
®
l
\ = Þl =l
l
17. (a) max
22
max
1 114
1213
3
(1) (2)
RÅ
R
éù
= - Þl =»
êú
l
êú
ëû
and min
2
min
1 111
910
(1)
RÅ
R
éù
= - Þl =»
êú
l¥
êú
ëû
18. (a) Maximum energy is liberated for transition 1
n
E ® and
minimum energy for 
1 nn
EE
-
®
Hence 
1
1
2
52.224
E
E eV
n
-=
.....(i)
and 
11
22
1.224
( 1)
EE
eV
nn
-=
-
......(ii)
Solving equations (i) and (ii) we get
1
54.4 E eV =- and 
5 n =
Now 
2
1
2
13.6
54.4
1
Z
E eV =- =-
.  Hence 
2 Z =
19. (d) Radius
22
0
2
nh
R
nZe
e
=
p
V elocity 
2
0
2
Ze
v
nh
=
e
and energy 
24
222
0
8
mZe
E
nh
=-
e
Now, it is clear from above expressions 
. Rvn µ
20. (c) At closest distance of approach
Kinetic energy = Potential energy
6 19
0
1 ( )(2)
5 10 1.6 10
4
Zee
r
-
Þ´ ´ ´ =´
pe
For uranium Z = 92, so 
12
5.3 10 r cm
-
=´
21. (a) Here radius of electron orbit 
1/ rm µ
and energy
Em µ
, where m is the mass of the electron.
Hence energy of hypothetical atom
0
2 ( 13.6 ) 27.2 E eV eV = ´ - =- and radius 
0
0
2
a
r =
22. (a) Time period, T
n
 = 
n
n
2r
v
p
 (in n
th
 state)
i.e. 
n
n
n
r
T
v
µ
 But 
2
n
rn µ and 
n
1
v
n
µ
Therefore, 
3
n
Tn µ .
Given 
nn
11
T 8T =
, Hence n
1
 = 2n
2
. Þ n
1
 is even
23. (d) 2.55eV = E
4
 – E
2
.
Therefore other photon will have energy
        = E
2
– E
1
 = 10.2 eV.
Energy given to H-atom excitation = E
2
 – E
1
 = 12.75 eV .
Consider perfectly inelastic collision for other answer.
24. (a) Balmer series lies in the visible region.
25. (b),     26. (d),       27.  (a)
Since 6 different types of photons are emitted implies 
4
C
2
i.e. highest excitation state is n = 4
Since emission energies are equal, lesser and greater so initial
state
2
12420 1 1
e 13.6Z
4 16
éù
= =-
êú
l
ëû
Þ Z
2
 = 16 Þ Z = 4
41
11
E 13.6 (16) 20.4eV
1 16
®
= -=
43
11
E 13.6 (16) 10.6 eV
9 16
®
= -=
28. (b) Bohr postulated that electrons in stationary orbits
around the nucleus do not radiate.
This is the one of Bohr’s postulate. According to this
the moving electrons radiate only when they go from
one orbit to the next lower orbit.
29. (b) Rutherford confirmed the repulsive force on a-particle
due to nucleus varies with distance according to inverse
square law and that the positive charges are
concentrated at the centre and not distributed
throughout the atom.
30. (b) When the atom gets appropriate energy from outside,
then this electron rises to some higher energy level.
Now it can return either directly to the lower energy
level or come to the lowest energy level after passing
through other lower energy lends, hence all possible
transitions take place in the source and many lines are
seen in the spectrum.