CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A metal surface is illuminated by light of two different wavelengths λ1=500 nm and λ2=620 nm. The maximum speed of ejected photoelectrons corresponding to these wavelengths are in the ratio 3:2. Then work function of the metal in milli eV is approximately equal to .
(hc=1240 eV – nm)

A
01.62
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
E
3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
F
3.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
G
1.6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
H
1.60
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
I
1616.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
J
1616
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
K
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
L
1616.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

Use equation of photoelectric effect
hcλ=ϕ+K.EMax
So, we have
given, hc=1240 eV-nm and v1v2=32
Therefore,
1240500=ϕ+12mv21 ..(i)
1240620=ϕ+12mv22 ..(ii)
On solving the above two equation we get
ϕ=1616 meV

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Work Function
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon