a) 2:3
b) 3:4
c) 1:3
d) √3: √2
[2020 | Delhi | Section - A | SET - 1 SET - 3, Q.8 Q.3]
Solution :
We know,
Kmax = hν-hν₀
Kmax =h(ν-ν₀)
Now
KA=h(ν-ν/2)
KA=hν/2
KB=h(ν-ν/3)
KB=h(2ν/3)
KB=2hν/3
KA/KB =[hν/2]/[2hν/3]
KA/KB=3/4
KA : KB = 3:4
041220 | Dual Nature of Radiation and Matter
If the photons of frequency ν are incident on the surfaces of metals. A & B of threshold frequencies ν/2 and ν/3 respectively, the ratio of the maximum kinetic energy of electrons emitted from A to that from B is
a) 2:3
b) 3:4
c) 1:3
d) √3: √2
[2020 | Delhi | Section - A | SET - 1 SET - 3, Q.8 Q.3]
a) 2:3
b) 3:4
c) 1:3
d) √3: √2
[2020 | Delhi | Section - A | SET - 1 SET - 3, Q.8 Q.3]
