what is kinetic energy of emitted electron when Caesium is exposed to UV rays of frequency 1.2×10^15 Hz?
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(a). Work function, ϕ
o
=2.14eV
Frequency of light, ν=6×10
14
Hz
Maximum K.E. is given by the photoelectric effect.
K=hν−ϕ
o
K=(
1.6×10
−19
6.626×10
−34
×6×10
14
)−2.14=0.345eV
(b). For stopping potential, V
o
, we can write the equation for kinetic energy as,
K=eV
o
⇒V
o
=
e
K
=
1.6×10
−19
0.345×1.6×10
−19
=0.345V
(c). The maximum speed of the emitted photoelectrons is v.
Hence, the relation for kinetic energy can be written as,
K=
2
1
mv
2
m=9.1×10
−31
kg is the mass of electron.
⇒v
2
=2K/m
⇒v=3.323×10
5
m/s=332.3km/s
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