Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 2.05×1014 s−1.?
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, 𝜈0 .
Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 2.05×1014 s−1.
Update:With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of 255 nm?
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E = h𝜈
E = 6.626X10^-34 Js X 2.06X10^14 s^-1 = 1.36X10^-19 J
Ephoton = hc/wavelength
Ephoton = (6.626X10^-34 Js)(3.00X10^8 m/s) / 2.55X10^-7 m = 7.79X10^-19 J
Ke = 7.99X10^-19 J - 1.36X10^-19 J = 6.43 X10^-19 J
The energy of a photon is h*nu, where nu is the frequency and h is Planck's constant, 6.626 x 10^(-34) J*s.
So the answer to your question is around 1.2 x 10^(-19) J but use a calculator.
you can use E=hf to find the minimum energy (which would be the work function of that metal)
and h is planck's constant
for the maximum kinetic energy you have max Kinetic energy=hf-work function
and remember that you can find frequency using the wavelength as frequency=wavespeed/wavelength and wavespeed of light is about 3x10^8