The work function for potassium and cesium are 2.25 and 2.14 eV, respectively.

Question

a) Will the photoelectric effect occur for either of these elements with incident light of wavelength 565 nm?
b) With light of wavelength 518 nm ?

Answer 1

convert 565 and 518 to wavelength.

Fnergy= hc/lambda

set lambda=hc/energy

now convert eV to joules, plug it in and compute lambda cutoff for each metal. If lambda oncoming is shorter than 565, and 518nm, photoelectric effect occurs.

Answer 2

W potassium eV 0 ( ) = 2.25 ,
W caesium eV 0 ( ) = 2.14 ,
7
1 5650 10 m
−
λ = × ,
7
2 5180 10 m
−
λ = ×
34
h J s 6.63 10 .
−
= × ,
8
c m s = × 3 10 /
W potassium eV 0 ( ) = 2.25
19
2.25 1.6 10 J
−
= × ×
19
3.6 10 J
−
= ×
W caesium eV 0 ( ) = 2.14
19
2.14 1.6 10 J
−
= × ×
19
3.424 10 J
−
= ×
i.
34 8
7
1
6.63 10 3 10
5.650 10
hc
−
−
× × ×
=
� ×

19
3.52 10 J
−
= ×
This energy (of the
incident photon) is greater
than W0
(caesium), but
less than W0
(potassium).
Hence, photoelectric effect
will occur in case of
caesium, but not in case
of potassium.
ii.
34 8
7
2
6.63 10 3 10
5.180 10
hc
−
−
× × ×
=
λ ×

19
3.84 10 J
−
= ×
This is greater than W0
for potassium and for
caesium. Hence,
photoelectric effect will occur in both cases.