Two identical, photocathodes receive light of frequencies f1 and f2. If the velocities of the photo electrons (of mass m) coming out are respectively v1 and v2, then (where the symbols have their usual meaning)
A
v21−v22=2hm(f1−f2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
v1+v2=(2hm(f1+f2))12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
v21+v22=2hm(f1+f2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
v1−v2=(2hm(f1−f2))12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Av21−v22=2hm(f1−f2) hf=hf0+12mv2Here,v21=2hf1m−2hf2m⇒v21−v22=2hm(f1−f2)