Respuesta :
Answer:
1 eV
Explanation:
Given:
Work function, ∅ = 2.00 eV
Kinetic energy of the ejected of the electron, K.E = 4.0 eV
Now,
using the photoelectric equation , we have
Energy of the photon (E) = ∅ + K.E
also,
E = hc/λ
where, h is plank's constant
c is the speed of the light
λ is the wavelength
thus, we have
hc/λ = 2 + 4 = 6 eV
Energy of photon = 6eV
Now,
for the second case
λ' = 2λ
when Wavelength is doubled , E is halved
thus,
E' = hc/λ'
or
E' = hc/2λ
or
E' = E/2 = 6/2 = 3 eV
also,
E' = ∅ + KE '
thus on substituting the values,
3 = 2 + KE'
or
KE' = 1 eV
Hence, the maximum kinetic energy for the second case is 1 eV
The maximum energy of the second source is approximately equal to zero
WORK FUNCTION
The work function of a metal is the minimum energy required to liberate an electron from a metal surface.
The Maximum kinetic energy of the electron can be expressed as
E = hf - W
Where
E = Total energy
hf = energy of the electron
w = work function
Given that a light source of wavelength λ illuminates a metal with a work function of W = 2.00 eV and ejects electrons with a maximum K.E = 4.00 eV.
E = hf - w
4 x 1.6 x [tex]10^{-19}[/tex] = hf - (2 x 1.6 x [tex]10^{-19}[/tex] )
hf = 3.2 x [tex]10^{-19}[/tex]
h = plank constant = 6.6 x [tex]10^{-34}[/tex] Js
f = 3.2 x [tex]10^{-19}[/tex] / 6.6 x [tex]10^{-34}[/tex]
f = 4.85 x [tex]10^{14}[/tex] Hz
By using the speed of light, we can calculate the wavelength. That is,
λ = V/F
λ = 3 x [tex]10^{8}[/tex] / 4.85 x [tex]10^{14}[/tex]
λ = 6.2 x [tex]10^{-7}[/tex] m
Given that a second light source with double the wavelength of the first ejects photoelectrons, then, the wavelength will be 2 x 6.2 x [tex]10^{-7}[/tex] m
The frequency can be calculated by using wave speed formula
F = V / λ
F = 3 x [tex]10^{8\\[/tex] / 6.2 x [tex]10^{-7}[/tex]
F = 4.84 x [tex]10^{14}[/tex] Hz
The energy of the incident photon will be
hf = 6.6 x [tex]10^{-34}[/tex] x 4.84 x [tex]10^{14}[/tex]
hf = 3.1944 x [tex]10^{-19}[/tex] J
The maximum kinetic energy of the second source will be
E = hf - w
E = 3.1944 x [tex]10^{-19}[/tex] - 3.2 x [tex]10^{-19}[/tex]
E = Approximately zero.
Therefore, the maximum energy of the second source is approximately equal to zero.
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