Dual charecter of electron

photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it. In a broader definition, the radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the material may be a solid, liquid, or gas; and the released particles may be ions (electrically charged atoms or molecules) as well as electrons. The phenomenon was fundamentally significant because of the puzzling questions it raised about the nature of light—particle versus wavelike behaviour—that were finally resolved by Albert Einstein in 1905. The effect remains important for research in areas from materials science to astrophysics, as well as forming the basis for a variety of useful devices.

Photoelectric principles

According to quantum mechanics, electrons bound to atoms occur in specific electronic configurations. The highest energy configuration (or energy band) that is normally occupied by electrons for a given material is known as the valence band, and the degree to which it is filled largely determines the material’s electrical conductivity. In a typical conductor (metal), the valence band is about half filled with electrons, which readily move from atom to atom, carrying a current. In a good insulator, such as glass or rubber, the valence band is filled, and these valence electrons have very little mobility. Like insulators, semiconductors generally have their valence bands filled, but, unlike insulators, very little energy is required to excite an electron from the valence band to the next allowed energy band—known as the conduction band, because any electron excited to this higher energy level is relatively free. For example, the “bandgap” for silicon is 1.12 eV (electron volts), and that of gallium arsenide is 1.42 eV. This is in the range of energy carried by photons of infrared and visible light, which can therefore raise electrons in semiconductors to the conduction band. (For comparison, an ordinary flashlight battery imparts 1.5 eV to each electron that passes through it. Much more energetic radiation is required to overcome the bandgap in insulators.) Depending on how the semiconducting material is configured, this radiation may enhance its electrical conductivity by adding to an electric current already induced by an applied voltage (see photoconductivity), or it may generate a voltage independently of any external voltage sources (see photovoltaic effect).

Photoconductivity arises from the electrons freed by the light and from a flow of positive charge as well. Electrons raised to the conduction band correspond to missing negative charges in the valence band, called “holes.” Both electrons and holes increase current flow when the semiconductor is illuminated.

F &Q

Solved Problems

Question 1: An electron and a photon have the same wavelength. If p is the momentum of the electron and E is the energy of the photon. The magnitude of p/E in S.I unit is

(a) 3.0108 (b) 3.3310-9

(c) 9.110-31 (d) 6.6410-34

Answer: b

As we know that, for electron, λ = h/p

Or

p = h/λ

And for photon E = hc / λ

Thus, p / E = 1 / c = 1 / (3 x 108 m/s) = 0. 33 x 10-8 s/m

Question 2: What is the energy and wavelength of a thermal neutron?

Answer:

KE = (3/2) kT = (3/2) (1.38 × 10–23) (293) = 607 × 10–23 J

λ = h / p

λ = h / √2m0 (KE)

λ = 6 .63 × 10–34 / √2 (1.67 × 10–27) (607 × 10–23)

λ = 0.147 nm

Question 3: A particle of mass m is confined to a narrow tube of length L. Find

(a) The wavelengths of the de-Broglie wave which will resonate in the tube,

(b) The corresponding particle momenta, and

(c) The corresponding energies.

Answer:

(a) The de Broglie waves will resonate with a node at each end of the tube.

A few of the possible resonance forms are listed below:

λn = (2L / n); = 1, 2, 3, …

(b) Since de-Broglie wavelengths are

λn = h / pn

∴ pn = h / λn = nh / 2L, n = 1, 2, 3, …

(c) (KE) n = pn2 / 2m = n2h2/8L2m, n = 1, 2, 3, …

Question 4: What is the wavelength of an electron moving at 5.31 x 106 m/sec?

Given: mass of electron = 9.11 x 10-31 kg h = 6.626 x 10-34 J·s

Answer:

de Broglie’s equation is

λ = h/mv

λ = 6.626 x 10-34 J·s/ 9.11 x 10-31 kg x 5.31 x 106 m/sec

λ = 6.626 x 10-34 J·s/4.84 x 10-24 kg·m/sec

λ = 1.37 x 10-10 m

λ = 1.37 Å

The wavelength of an electron moving 5.31 x 106 m/sec is 1.37 x 10-10 m or 1.37 Å.

Question 5: Which of the following is called non-mechanical waves?

a) Magnetic waves

b) Electromagnetic waves

c) Electrical waves

d) Matter waves

Answer: b

The waves which travel in the form of oscillating electric and magnetic waves are called electromagnetic waves. Such waves do not require any material for their propagation and are called non-mechanical waves.

Question 6: Which of the following is associated with an electron microscope?

a) Matter waves

b) Electrical waves

c) Magnetic waves

d) Electromagnetic waves

Answer: a

The waves associated with microscopic particles when they are in motion are called matter waves. Electron microscope makes use of the matter waves associated with fast-moving electrons.

Question 7: Calculate the de-Broglie wavelength of an electron that has been accelerated from rest on application of a potential of 400volts.

a) 0.1653 Å

b) 0.5125 Å

c) 0.6135 Å

d) 0.2514 Å

Answer: c

de-Broglie wavelength = h/√ (2×m×e×V)

de-Broglie wavelength = (6.625×10-14)/√(2×9.11×10-31×1.6×10-17×400)

Wavelength = 0.6135 Å.

Question 8: The de Broglie wavelength of a particle is the same as the wavelength of a photon. Then, the photon’s energy is:

(a) Equal to the kinetic energy of the particle.

(b) less than the kinetic energy of the particle.

(c) greater than the kinetic energy of the particle.

(d) Nothing can be specified.

Answer: c

The photon’s energy depends only on the frequency of the photon. Hence, the photon’s energy is greater than the kinetic energy of the electron.

Question. 9: An electron and a proton have the same de Broglie wavelength. Then the kinetic energy of the electron is?

(a) Zero

(b) Infinity

(c) Equal to the kinetic energy of the proton

(d) Greater than the kinetic energy of the proton

Answer: d

The electron and the proton have the same de Broglie wavelength, which means their momenta are the same. Since the mass of the proton is greater than the mass of the electron, its speed is less than the speed of the electron. Hence, KE = (½) PV, the kinetic energy of the electron is greater than the kinetic energy of the proton.

Question. 10: A nucleus of mass M at rest emits an α-particle of mass m. The de Broglie wavelengths of the α-particle and residual nucleus will be in the ratio

(a) m : M (b) (M+m) : m

(c) M : m (d) 1 : 1

Answer: d

The nucleus is initially at rest. By the conservation of momentum principle, we can assume that the momenta of the α-particle of mass m and the residual nucleus will be equal and opposite. The de Broglie wavelength is inversely proportional to momentum. Hence, the de Broglie wavelengths of the α-particle and residual nucleus will be in the ratio = 1:1.

Question 11: The operation of the electron microscope depends on the:

(a) The very short wavelength of X-rays

(b) Wave nature of electrons

(c) Electromagnetic theory of waves

(d) Photoelectric effect

Answer: b

The wave nature of the electron particles is utilized in the electron microscope.