Advertisements
Advertisements
Question
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?
Advertisements
Solution
The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
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?
The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
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?
In an experiment on photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in the opposite direction. Does this violate the principle of conservation of momentum?
Planck's constant has the same dimensions as
Let nr and nb be the number of photons emitted by a red bulb and a blue bulb, respectively, of equal power in a given time.
The work function of a metal is hv0. Light of frequency v falls on this metal. Photoelectric effect will take place only if
A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential
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
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)
Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 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)
When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
Answer the following question.
Plot a graph of photocurrent versus anode potential for radiation of frequency ν and intensities I1 and I2 (I1 < I2).
In the case of photoelectric effect experiment, explain the following facts, giving reasons.
The photoelectric current increases with increase of intensity of incident light.
Explain how does (i) photoelectric current and (ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same?
Show the graphical variation in the above two cases.
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
- What does X and A on the horizontal axis represent?
- Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3 (ʋ3 > ʋ2 > ʋ1) for the same intensity.
- Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3 > I2 > I1) having the same frequency.
If photons of ultraviolet light of energy 12 eV are incident on a metal surface of work function of 4 eV, then the stopping potential (in eV) will be :
