Question
Question: If \[{{\lambda }_{0}}\] and \[\lambda \] be the threshold wavelength and wavelength of incident ligh...
If λ0 and λ be the threshold wavelength and wavelength of incident light, the
velocity of photoelectron ejected from the metal surface is
(a) m2h(λ0−λ)
(b) m2hc(λ0−λ)
(c) m2h(λλ0λ0−λ)
(d) m2h(λ01−λ1)
Solution
Hint: Threshold wavelength is defined as the minimum wavelength of incident light required in order to eject the electrons from the metal surface. Once the threshold is crossed the metal surface exhibits the photoelectric effect.
Complete step by step solution:
We already know that,
hv=ϕ+KineticEnergy (i)
where, v=incident frequency
Planck’s constant (h) = 6.62×10−34Jsec
ϕ=threshold energy
We have been given in the question that
ϕ=λ0hc (ii)
The equation of threshold energy is given as
hv=λhc (iii)
And for Kinetic energy is given as,
K.E.=21mVmax2 (iv)
From equation (ii)(iii)(iv) and (i), we get,
λhc=λ0hc+21mVmax2
On rearranging,
λhc−λ0hc=21mVmax2
We will take hc as common,
hc(λ1−λ01)=21mVmax2
Again, on rearranging,
m2hc(λ1−λ01)=Vmax2
Finally, we get,
Vmax2=m2h(λλ0λ0−λ)
So, the correct option is (c).
Additional Information:
The photoelectric effect can be explained by assuming that radiation consists of particles of light called photons and each photon carries a quantum of energy. The energy of a photon depends solely on its frequency, which is the frequency of the radiation
The maximum kinetic energy of a photoelectron is the difference between the energy of the incident photon and the work function of the metal, where the work function is the binding energy of electrons to the metal surface. Each metal has its own characteristic work function.
Note: The name threshold wavelength differs from literature to literature. Don’t get confused between threshold wavelength, threshold frequency or work function. Essentially all have the same meaning.