Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exams
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
State Eligibility Tests
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
English
Competitive English
Board Exams
Andhra
CBSE
ICSE
Kerala
Maharashtra
Tamil Nadu
Telangana
Uttar Pradesh
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
IIT JEE Advanced 2009 Question Paper 2
Show Para
Hide Para
Share question:
© examsnet.com
Question : 2
Total: 57
Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions
ϕ
p
= 2.0 eV,
ϕ
q
= 2.5 eV and
ϕ
r
= 3.0 eV, q r respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I–V graph for the experiment is (Take hc = 1240 eV nm)
Validate
Solution:
We have
E =
(
h
c
λ
)
J ⇒ E =
(
1240
λ
n
m
)
eV
Therefore,
λ
1
= 550 nm,
E
1
= 2.25 eV
λ
2
= 450 nm ,
E
2
= 2.75 eV
λ
3
= 350 nm ,
E
3
= 3.5 eV
Also,
ϕ
p
= 2 eV, all λ’s cause emissions.
ϕ
q
= 2.5 eV, last two λ’s cause emissions.
ϕ
r
= 3 eV, only the last λ causes emissions.
That is,
I
p
>
I
q
>
I
r
.
© 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
Prev Question
Next Question