find the velocity and kinetic energy of electron having wavelength 600 NM
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Answer:
Here, λ=600nm,λ
′
=400nm,K
max
′
=2K
max
According to Einstein photoelectric equation
K
max
=
λ
hc
−ϕ
0
and 2K
max
=
λ
′
hc
−ϕ
0
Dividing by we get
2=
λ
hc
−ϕ
0
λ
′
hc
−ϕ
0
or
λ
2hc
−2ϕ
0
=
λ
′
hc
−ϕ
0
λ
2hc
−2ϕ
0
=
λ
′
hc
−ϕ
0
Or hc(
λ
2
−
λ
′
1
)=ϕ
0
∴ϕ
0
=1240eVnm(
600nm
2
−
400nm
1
)=1.03eV
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