Atoms Practice Questions - DPP for JEE

 Page 1


1. (c) As 
and 
?  
or, 
or, 
For n
th
 orbit
?  r
n
 = a
0
n
2
 – ß
2. (c) Magnetic moment of the hydrogen atom, when the electron is in 
n
th
 excited state, i.e., n' = (n + 1)
As magnetic moment M
n
 = I
n
A = i
n
(pr
n
? 2
)
Solving we get magnetic moment of the hydrogen atom for n
th
 excited
state
3. (c) ; 
Page 2


1. (c) As 
and 
?  
or, 
or, 
For n
th
 orbit
?  r
n
 = a
0
n
2
 – ß
2. (c) Magnetic moment of the hydrogen atom, when the electron is in 
n
th
 excited state, i.e., n' = (n + 1)
As magnetic moment M
n
 = I
n
A = i
n
(pr
n
? 2
)
Solving we get magnetic moment of the hydrogen atom for n
th
 excited
state
3. (c) ; 
or     
or 
or 
4. (a) Number of emission spectral lines
 in first case.
Or  
Take positive root.
? n
1
 = 3
Again,  in second case.
Or 
Take positive root, or n
2
 = 4
Now velocity of electron 
5. (b) E = – 3.4 eV and 
angular momentum = mvr
m
2
v
2
= 99.008 × 10
–50
mv = 9.95028 × 10
–25
Page 3


1. (c) As 
and 
?  
or, 
or, 
For n
th
 orbit
?  r
n
 = a
0
n
2
 – ß
2. (c) Magnetic moment of the hydrogen atom, when the electron is in 
n
th
 excited state, i.e., n' = (n + 1)
As magnetic moment M
n
 = I
n
A = i
n
(pr
n
? 2
)
Solving we get magnetic moment of the hydrogen atom for n
th
 excited
state
3. (c) ; 
or     
or 
or 
4. (a) Number of emission spectral lines
 in first case.
Or  
Take positive root.
? n
1
 = 3
Again,  in second case.
Or 
Take positive root, or n
2
 = 4
Now velocity of electron 
5. (b) E = – 3.4 eV and 
angular momentum = mvr
m
2
v
2
= 99.008 × 10
–50
mv = 9.95028 × 10
–25
L = (9.95028 × 10
–25
) 
= 2.10 × 10
–34
 Js.
6. (b) Radius of circular path followed by electron is given by,
?
For transition between 3 to 2.
Work function = 1.88 eV – 0.8 eV = 1.08 eV ˜ 1.1eV
7. (d) Radius of n
th
 orbit r
n
 ? n
2
, graph between r
n
 and n is a parabola.
Also, 
Comparing this equation with y = mx + c,
Graph between  and log
e
 (n) will be a straight line, passing
from origin.
Similarly it can be proved that graph between  and log
e
 (n) is
a straight line. But with negative slops.
8. (a)
= 993 A°
(where Rydberg constant , R = 1.097 × 10
7
)
Page 4


1. (c) As 
and 
?  
or, 
or, 
For n
th
 orbit
?  r
n
 = a
0
n
2
 – ß
2. (c) Magnetic moment of the hydrogen atom, when the electron is in 
n
th
 excited state, i.e., n' = (n + 1)
As magnetic moment M
n
 = I
n
A = i
n
(pr
n
? 2
)
Solving we get magnetic moment of the hydrogen atom for n
th
 excited
state
3. (c) ; 
or     
or 
or 
4. (a) Number of emission spectral lines
 in first case.
Or  
Take positive root.
? n
1
 = 3
Again,  in second case.
Or 
Take positive root, or n
2
 = 4
Now velocity of electron 
5. (b) E = – 3.4 eV and 
angular momentum = mvr
m
2
v
2
= 99.008 × 10
–50
mv = 9.95028 × 10
–25
L = (9.95028 × 10
–25
) 
= 2.10 × 10
–34
 Js.
6. (b) Radius of circular path followed by electron is given by,
?
For transition between 3 to 2.
Work function = 1.88 eV – 0.8 eV = 1.08 eV ˜ 1.1eV
7. (d) Radius of n
th
 orbit r
n
 ? n
2
, graph between r
n
 and n is a parabola.
Also, 
Comparing this equation with y = mx + c,
Graph between  and log
e
 (n) will be a straight line, passing
from origin.
Similarly it can be proved that graph between  and log
e
 (n) is
a straight line. But with negative slops.
8. (a)
= 993 A°
(where Rydberg constant , R = 1.097 × 10
7
)
or,  
Solving we get n
2
 = 3
Spectral lines
Total number of spectral lines = 3
Two lines in Lyman series for n
1
 = 1, n
2
 = 2 and n
1
 = 1, n
2
 = 3 and one in
Balmer series for n
1
 = 2 , n
2
 = 3
9. (a) Given potential energy between electron  and proton
= eV
0
 log 
?    | F | 
But this force acts as centripetal force
?  ...(i)
By Bohr’s postulate, mvr ....(ii)
From (i) and (ii),
  ? r ? n
For n
th
 orbit r
n
 ? n
Page 5


1. (c) As 
and 
?  
or, 
or, 
For n
th
 orbit
?  r
n
 = a
0
n
2
 – ß
2. (c) Magnetic moment of the hydrogen atom, when the electron is in 
n
th
 excited state, i.e., n' = (n + 1)
As magnetic moment M
n
 = I
n
A = i
n
(pr
n
? 2
)
Solving we get magnetic moment of the hydrogen atom for n
th
 excited
state
3. (c) ; 
or     
or 
or 
4. (a) Number of emission spectral lines
 in first case.
Or  
Take positive root.
? n
1
 = 3
Again,  in second case.
Or 
Take positive root, or n
2
 = 4
Now velocity of electron 
5. (b) E = – 3.4 eV and 
angular momentum = mvr
m
2
v
2
= 99.008 × 10
–50
mv = 9.95028 × 10
–25
L = (9.95028 × 10
–25
) 
= 2.10 × 10
–34
 Js.
6. (b) Radius of circular path followed by electron is given by,
?
For transition between 3 to 2.
Work function = 1.88 eV – 0.8 eV = 1.08 eV ˜ 1.1eV
7. (d) Radius of n
th
 orbit r
n
 ? n
2
, graph between r
n
 and n is a parabola.
Also, 
Comparing this equation with y = mx + c,
Graph between  and log
e
 (n) will be a straight line, passing
from origin.
Similarly it can be proved that graph between  and log
e
 (n) is
a straight line. But with negative slops.
8. (a)
= 993 A°
(where Rydberg constant , R = 1.097 × 10
7
)
or,  
Solving we get n
2
 = 3
Spectral lines
Total number of spectral lines = 3
Two lines in Lyman series for n
1
 = 1, n
2
 = 2 and n
1
 = 1, n
2
 = 3 and one in
Balmer series for n
1
 = 2 , n
2
 = 3
9. (a) Given potential energy between electron  and proton
= eV
0
 log 
?    | F | 
But this force acts as centripetal force
?  ...(i)
By Bohr’s postulate, mvr ....(ii)
From (i) and (ii),
  ? r ? n
For n
th
 orbit r
n
 ? n
10. (c)
The wave number  of the radiation = 
= 
Now for case (I)   n
1 
= 3, n
2 
= 2
, R
8
 =  Rydberg constant
? 
?
11. (b) E =