Respuesta :
From Literature:
The amount of energy in the photons is given by this equation:
E = hf
where E = energy
h = Planck's constant = 6.63 * 10^-34 Joule seconds
f = frequency of the light, Hz
Given:
E= 3.00 eV and Planck's constant
To solve for the frequency, E = 3.00 eV
1 electronvolt = 1.60218 x 10^-19 Joules
3 * 1.60218 x 10^-19 Joules = 6.63 * 10^-34 Joule seconds * f
f = 7.25 x 10^14 /second or hertz
Therefore, the threshold frequency of the material is 7.25 x 10^14 Hertz.
The amount of energy in the photons is given by this equation:
E = hf
where E = energy
h = Planck's constant = 6.63 * 10^-34 Joule seconds
f = frequency of the light, Hz
Given:
E= 3.00 eV and Planck's constant
To solve for the frequency, E = 3.00 eV
1 electronvolt = 1.60218 x 10^-19 Joules
3 * 1.60218 x 10^-19 Joules = 6.63 * 10^-34 Joule seconds * f
f = 7.25 x 10^14 /second or hertz
Therefore, the threshold frequency of the material is 7.25 x 10^14 Hertz.
Answer:
The threshold frequency of this material is [tex]4.83\times 10^{14}\ Hz[/tex]
Explanation:
It is given that,
Energy of incident photon, E = hυ = 5 eV
Maximum kinetic energy of electrons, [tex]E_k=3\ eV[/tex]
We have to find the threshold frequency of this material. Einstein photoelectric equation gives the relationship between energy and threshold frequency i.e.
[tex]h\nu=h\nu_0+E_k[/tex]............(1)
where
[tex]\nu_0[/tex] is the threshold frequency of this material.
h is Planck's constant, [tex]h=4.1357\times 10^{-15}\ eV\ s[/tex]
Putting all values in equation (1) as :
[tex]5\ eV=h\nu_0+3\ eV[/tex]
[tex]\nu_0=\dfrac{2\ eV}{4.1357\times 10^{-15}\ eV\ s}[/tex]
[tex]\nu_0=4.83\times 10^{14}\ Hz[/tex]
Hence, the threshold frequency of this material is [tex]4.83\times 10^{14}\ Hz[/tex]