The work function for potassium and cesium are 2.25 and 2.14 eV, respectively.
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 ?
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.
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.
To determine whether the photoelectric effect will occur for potassium and cesium with incident light of different wavelengths, we need to compare the energy of the incident photons (determined by their wavelength) with the work function of each element.
The energy of a photon can be calculated using the equation: E = hc/λ, where E is energy, h is Planck's constant (approximately 6.63 x 10^-34 J⋅s), c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength of light in meters.
a) For incident light with a wavelength of 565 nm (565 x 10^-9 meters):
Calculating the energy of the incident photons:
E = (6.63 x 10^-34 J⋅s)(3.00 x 10^8 m/s) / (565 x 10^-9 m)
E ≈ 3.35 x 10^-19 J
Comparing the energy of the incident photons to the work function of potassium (2.25 eV):
1 eV ≈ 1.60 x 10^-19 J
2.25 eV ≈ 2.25 x 1.60 x 10^-19 J = 3.60 x 10^-19 J
Since the energy of the incident photons (3.35 x 10^-19 J) is less than the work function of potassium (3.60 x 10^-19 J), the photoelectric effect will not occur for potassium with incident light of wavelength 565 nm.
Similarly, for cesium with incident light of wavelength 565 nm:
Comparing the energy of the incident photons to the work function of cesium (2.14 eV):
2.14 eV ≈ 2.14 x 1.60 x 10^-19 J = 3.42 x 10^-19 J
Since the energy of the incident photons (3.35 x 10^-19 J) is greater than the work function of cesium (3.42 x 10^-19 J), the photoelectric effect will occur for cesium with incident light of wavelength 565 nm.
b) Similarly, let's perform the calculations for light with a wavelength of 518 nm:
Calculating the energy of the incident photons:
E = (6.63 x 10^-34 J⋅s)(3.00 x 10^8 m/s) / (518 x 10^-9 m)
E ≈ 3.82 x 10^-19 J
Comparing the energy of the incident photons to the work functions of potassium and cesium:
- For potassium (2.25 eV):
2.25 eV ≈ 2.25 x 1.60 x 10^-19 J = 3.60 x 10^-19 J
Since the energy of the incident photons (3.82 x 10^-19 J) is greater than the work function of potassium (3.60 x 10^-19 J), the photoelectric effect will occur for potassium with incident light of wavelength 518 nm.
- For cesium (2.14 eV):
2.14 eV ≈ 2.14 x 1.60 x 10^-19 J = 3.42 x 10^-19 J
Since the energy of the incident photons (3.82 x 10^-19 J) is also greater than the work function of cesium (3.42 x 10^-19 J), the photoelectric effect will occur for cesium with incident light of wavelength 518 nm.
In conclusion:
a) The photoelectric effect will occur for cesium, but not for potassium, with light of wavelength 565 nm.
b) The photoelectric effect will occur for both potassium and cesium with light of wavelength 518 nm.