The light of wavelength 4000 angstrom falls on a metal whose critical wavelength of photoelectric effect is 6600 angstrom .Calculate the kinetic energy .
The light of wavelength 4000 angstrom falls on a metal whose critical wavelength of photoelectric effect is 6600 angstrom .Calculate the kinetic energy .
The frequency and energy of the photon falling on the surface of the metal would be the same as the frequency of radiation. The energy of the photon will be
E = hν
= (hXc)/λ
where ν is the frequency of light, λ is the wavelength of light, c is the speed of light and h is the planck’s constant
This energy can be written as
E = hν0 + KE
where hν0 is the threshold frequency (minimum energy required to remove an electron from the surface of the metal and KE is the kinetic energy of the ejected electron.
Therefore
KE = E - hν0
= hν - hν0
= [(hXc)/λ - (hXc)/λ0]
= h X c [(1/ λ) – (1)/λ0)]
= 6.626 X 10-34 X 3 X 108 [{1/ (4000 X 10-10)}- {1/(6600 X 10-10)}]
= 1.6565 X 10-39 J
The frequency and energy of the photon falling on the surface of the metal would be the same as the frequency of radiation. The energy of the photon will be
E = hν
= (hXc)/λ
where ν is the frequency of light, λ is the wavelength of light, c is the speed of light and h is the planck’s constant
This energy can be written as
E = hν0 + KE
where hν0 is the threshold frequency (minimum energy required to remove an electron from the surface of the metal and KE is the kinetic energy of the ejected electron.
Therefore
KE = E - hν0
= hν - hν0
= [(hXc)/λ - (hXc)/λ0]
= h X c [(1/ λ) – (1)/λ0)]
= 6.626 X 10-34 X 3 X 108 [{1/ (4000 X 10-10)}- {1/(6600 X 10-10)}]
= 1.6565 X 10-39 J
