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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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.

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प्रश्न

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.]

संख्यात्मक
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उत्तर

Einstein’s photoelectric equation is given as:

eV0 = hv − `phi_0`

`"V"_0 = "h"/"e" "v" - phi_0/"e"` .............(1)

Where,

V0 = Stopping potential

h = Planck’s constant

e = Charge on an electron

v = Frequency of radiation

`phi_0` = Work function of a material

It can be concluded from equation (1) that potential V0 is directly proportional to frequency v.

Frequency is also given by the relation:

`"v" = "Speed of light (c)"/"Wavelenght (λ)"`

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

`"v"_1 = "c"/lambda_1 = (3 xx 10^8)/(3650 xx 10^(-10)) = 8.219 xx 10^14  "Hz"`

`"v"_2 = "c"/lambda_2 = (3 xx10^8)/(4047 xx 10^(-10)) = 7.412 xx 10^14  "Hz"`

`"v"_3 = "c"/lambda_3 = (3 xx 10^8)/(4358 xx 10^(-10)) = 6.884 xx 10^14  "Hz"`

`"v"_4 = "c"/lambda_4 = (3 xx 10^8)/(5461 xx 10^(-10)) = 5.493 xx 10^14  "Hz"`

`"v"_5 = "c"/lambda_5  = (3xx10^8)/(6907 xx 10^(-10)) = 4.343 xx 10^14  "Hz"`

The given quantities can be listed in tabular form as:

Frequency × 1014 Hz 8.219 7.412 6.884 5.493 4.343
Stopping potential V0 1.28 0.95 0.74 0.16 0

The following figure shows a graph between νand V0.

It can be observed that the obtained curve is a straight line. It intersects the ν-axis at 5 × 1014 Hz, which is the threshold frequency (v0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the λ5 line, and therefore, no stopping voltage is required to stop the current.

 Slope of the straight line = `"AB"/"CB" = (1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

From equation (1), the slope `"h"/"e"` can be written as:

`"h"/"e" = (1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

∴ `"h" = (1.12 xx 1.6 xx 10^(-19))/(2.726 xx 10^(14))`

= 6.573 × 10−34 Js

The work function of the metal is given as:

`phi_0` = hv0

= 6.573 × 10−34 × 5 × 1014

= 3.286 × 10−19 J

= `(3.286 xx 10^(-19))/(1.6 xx 1^(-18))`

= 2.054 eV

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पाठ 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४१०]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 11 Dual Nature of Radiation and Matter
Exercise | Q 11.28 | पृष्ठ ४१०

संबंधित प्रश्‍न

Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?


The following graph shows the variation of photocurrent for a photosensitive metal : 


(a) Identify the variable X on the horizontal axis.

(b) What does the point A on the horizontal axis represent?

(c) Draw this graph for three different values of frequencies of incident radiation v1, v2 and v3 (v1 > v2 > v3) for same intensity.

(d) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I1 > I2 > I3) having same frequency.


Can a photon be deflected by an electric field? Or by a magnetic field?


Light of wavelength λ falls on a metal with work-function hc/λ0. Photoelectric effect will take place only if


If the frequency of light in a photoelectric experiment is doubled, the stopping potential will ______.


The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. A light source is put on and a saturation photocurrent is recorded. An electric field is switched on that has a vertically downward direction.


When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W m−2 of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5 × 1011 m. (a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. (b) How many photons are there in each cubic metre near the earth's surface at any instant? (c) How many photons does the sun emit per second?

(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 parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 × 1019. Calculate the force exerted by the light beam on the mirror.

(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)


The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

(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)


Consider a thin target (10–2 cm square, 10–3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100W/m2 (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m3].


The graph shows the variation of photocurrent for a photosensitive metal

  1. What does X and A on the horizontal axis represent?
  2. Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3 (ʋ3 > ʋ2 > ʋ1) for the same intensity.
  3. Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency.

Which part is the photosensitive metal plate?


Which apparatus applies a potential difference between the plates?


Which apparatus measures the potential difference?


How does collector plate A help measure current?


What type of potential can the battery create between the plates?


What happens when the collector is made negative with respect to the emitter?


What happens if the frequency of incident radiation is below the threshold frequency?


Which observation was one of the major difficulties for classical wave theory?


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