Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
JEE Mains 03-Sep-2020 Shift 1 Solved Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 48
Total: 75
The photoelectric current from Na (work function,
w
0
=
2.3
e
V
) is stopped by the output voltage of the cell
P
t
(
s
)
|
H
2
(
g
,
1
b
a
r
)
|
H
C
l
(
a
q
.
p
H
=
1
)
|
A
g
C
l
(
s
)
|
A
g
(
s
)
The pH of aq. HCl required to stop thephotoelectric current from
K
(
w
0
=
2.25
e
V
)
,
all other conditions remaining the same, is ______.
×
10
−
2
(to the nearest integer).
Given,
2.303
R
T
F
=
0.06
V
;
E
°
A
g
C
l
|
A
g
|
C
l
−
=
0.22
V
Your Answer:
Validate
Solution:
1
2
H
2
→
H
+
+
e
Θ
e
Θ
+
A
g
C
l
(
s
)
→
A
g
(
s
)
+
C
l
Θ
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
─
1
2
H
2
+
A
g
C
l
(
s
)
→
H
(
a
q
)
+
+
A
g
(
s
)
+
C
l
(
a
q
)
Θ
E
=
ε
0
−
.06
1
log
[
H
+
]
[
C
l
Θ
]
P
H
2
1
2
E
=
0.22
−
.06
log
(
10
−
1
)
(
10
−
1
)
1
1
2
E
=
0.22
+
.12
=
.34
v
o
l
t
⇒
total energy of photon will be (for Na)
=
2.3
+
0.34
=
2.64
e
V
⇒
stopping potential required for
K
=
2.64
−
2.25
=
0.39
volt
E
=
ε
0
−
.06
1
log
[
H
+
]
[
C
l
−
]
P
H
2
1
2
as
[
H
+
]
=
[
C
l
Θ
]
so
0.39
=
0.22
−
.06
log
[
H
+
]
2
1
1
2
0.17
=
+
.12
p
H
p
H
=
1.4166
⇒
1.42
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
Prev Question
Next Question
Examsnet