Illuminating a surface alternately with
wavelengths and when
it
is observed that corresponding
maximum velocities of electrons differ
by a factor Then the work function
of the metal is
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0
Answer:
η=
(v
2
)
max
(v
1
)
max
=
λ
2
1
−
λ
o
1
λ
1
1
−
λ
o
1
Thus, n
2
(
λ
2
1
−
λ
o
1
)=
λ
1
1
−
λ
o
1
⟹
λ
o
1
=
(n
2
−1)
(
λ
2
n
2
−
λ
1
1
)
So, A=
λ
o
2πhc
=
λ
2
2πhc
(n
2
−1)
(n
2
−
λ
1
λ
2
)
=1.88eV
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