wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If a photon of energy 7.08×1019J falls on lithium metal having work function 2.42eV then the kinetic energy of the ejected election will be:

A
1.5eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2.0eV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3.0eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.5eV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 2.0eV
Photoelectric equation is given by,
K.E=hνw
Where w=work function, hν=energy of the photon
hν=7.08×1019J=4.43eV, w=2.42eV
Now substituting the values ,
K.E=4.43eV2.42eV=2.01eV
K.E=2eV

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Photoelectric Effect
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon