Advertisements
Advertisements
Question
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.
Solution
Work function (or threshold energy) (W0): The minimum energy of incident radiation required to eject the electrons from the metallic surface is defined as the work function of that surface.
W0 = hv0 = `(hc)/λ_0` Joules; v0 = Threshold frequency; λ0 = Threshold wavelength
Work function in electron volt, W0(eV) = `(hc)/(eλ_0) = 12375/(λ_0(Å))`
Einstein's photoelectric equation is E = W0 + Kmax
Maximum energy = `hv - phi`
According to the problem for the first condition wavelength of light λ = 600 nm and for the second condition, the wavelength of light λ' = 400 nm
Also, the maximum kinetic energy for the second condition is equal to twice the kinetic energy in the first condition.
i.e., K'max = 2Kmax
then K'max = `(hc)/λ - phi`
⇒ 2Kmax = `(hc)/λ^' - phi`
⇒ `2(1230/600 - phi) = (1230/400 - phi)` ......[∵ hc = 1240 eV nm]
⇒ `phi = 1230/1230` = 1.02 eV
APPEARS IN
RELATED QUESTIONS
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10−19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
If an electron has a wavelength, does it also have a colour?
Photoelectric effect supports quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) electric charge of the photoelectrons is quantised
A photon of energy hv is absorbed by a free electron of a metal with work-function hv − φ.
The electric field associated with a light wave is given by `E = E_0 sin [(1.57 xx 10^7 "m"^-1)(x - ct)]`. Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10−9 C m−2 on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Define the term: stopping potential in the photoelectric effect.
In photoelectric effect the photo current ______.
