CLASS 12 DUAL NATURE OF RADIATION & MATTER
DUAL NATURE OF RADIATION & MATTER
✤ PHOTONS : These are the packets of energy.
Some terms related to photons:
1. Energy associated with a photon E=h𝜈 joules
Here h = 6.6 x 10^-34 JxS.
𝜈=frequency
2. speed of Photon is about 3x10^8 ms-1.
3. Photon can't be deflected by electric or magnetic field.
4. Photons always travel in straight path.
✤ Free electrons : In metals, electrons in outermost shell of atoms are
loosely bound. Such loosely bound electron are called free electrons.
✤ Work function : It is defined as the minimum energy required by an
electron to just escape from metal surface so as to overcome the
restraining forces at the surface.: W=h𝜈0
❆ Here 𝜈0-threshold unit is joule
frequency. or ev
ev = 1.6 x 10-19 J.
✤ Electron emission : The phenomenon of emission of electron from the
surface of metal is called electron emission.
✤ Method of emission: 1. Thermonic emission.
2. Photoelectric emission
3. Field emission
4. Secondary emission.
✤ Photoelectric emission: It is the phenomenon of emission of electron
from the surface of metals, when radiations of suitable frequency fall
on them. Ex - Alkali metals
✤ Threshold frequency: It is the minimum freq. of incident radiations of
required to emit the electrons from emitter.
§ at threshold freq. k. E of electron = 0
✤ Effect of intensity on photo electric current:
Current α intensity
(no of energy quanta
per unit area per unit time )
✤ Effect of Potential on current:
at stopping Potential
Kman = eVo
= ½mv²
✤ No. of electrons ejected from metal surface
For a particular frequency of incident radiation
the minimum(-ve) retarding potential Vo given to
plate for which the photoelectric current become
zero is called Stopping potential.
Effect of frequency
1. The value of stopping
Potential is different
for different frequency.
Stopping potential is
independent of Intensity.
2. More frequency, more stopping
Potential.
3. The value of saturated current depend upon
intensity of light. Intensity α 1/(Distance)2
Graphical relation b/w stopping Potential (Vo) and frequency
1. For a given metal plate, stopping Potential α frequency
2. at threshold frequency
Vo = 0
3. higher the work function,
greater frequency!
4. work function = ex magnitude of slope.
✤ Laws of Photo electric emission:
1. For a given metal and frequency of incident radiation, the number of
photoelectrons ejected per second is directly proportional to intensity of
incident radiation.
2. For a given metal, there is a minimum freq. required below which
emission of electrons does not take place, the freq. is called threshold
frequency.
3. K.E. of ejected photo electron does not depend upon the intensity. It
depend upon the freq. of incident radiation.
4. It is an instantaneous Process.
✤ EINSTEIN'S Equation
When energy is supplied to a photosensitive plate then some part of
energy is used as work function and remaining part is used to provide
K.E.
i.e. hv = w + k.e
hv = hv₀ + ½mv²
h(v-v₀) = ½mv²
✤ De-Broglie hypothesis:
→ According to de-broglie a moving material particle sometimes act as
a wave and some time as a particle and the wave associated with
particle control it in all respect. Wavelength of the wave
λ = h/P
λ = h/mv Here p = momentum
p = mv.
p = mc.
de broglie's wave length associated with moving electron:
Since we know K.E = 1/2 mv²
and K.E = eV.
→ 1/2 mv² = eV
v = √2eV/m
λ = h/mv.
λ = h/√2 mev
λ = 12.27/√V A°
Revision
1. Energy of a photon, E = hv
E = RC/λ
2. no. of photons ejected per second ∝ N = P/E
3. momentum of photon p = mc
p = h/λ
p = hv/c
4. work function w = hv₀ = hc/λ₀
5. K.E = 1/2 mv²max
= eV₀, V₀ → stopping potential
6. Einstein's Eq. h(v - v₀) = 1/2 mv²
7. Intensity of radiation = Energy / (area × time) = Power / area.
ev = 1.6 x 10⁻¹⁹ J
h = 6.6 x 10⁻³⁴ JxS.
8. K.E = p²/2m or momentum = √2mk
9. debroglie wave length λ= h/p = h/√2mk
10. wave length of wave associated with moving electron
λ = 12.27 Å/√V → voltage
11. wave length with temp
λ ∝ 1/√T T is kelvin
Some Imp: Questions:
1. For a monochromatic radiation incident on a photosensitive surface,
why do all photoelectrons not come out with same energy?
Sol. Different electrons belong to different energy levels in the conduction
band. They need different energies to come out the metal surface for
the same incident radiation.
2. How can you explain the emission of electron from metal by Einstein's
Equation?
Sol. By Einstein's h = Wo + Kmax, when a photon of energy hv is incident
on a metal, a part of this energy equal to work function Wo is used to
removing the electron and remaining energy is used as K.E.
3. (i) Metal A has higher work function and
higher threshold frequency.
(ii) Slope of Vo-v shows the ratio of
W0/e = hv0/e
(iii) Slope is same for all metals
4. Draw a plot showing variation of Photo electric current with collector
potential for two different frequency v1>v2. In which case stopping
potential is higher? Justify your Answer.
For v1 > v2
V0 > V2
5. Plot a graph showing variation of de broglie wave length λ versus 1/√v,
Here V is accelerating Potential for two Particle A and B carrying same
charge but masses m₁, m₂ (m₁>m₂)
Sol.
6. A proton and an-α-particle have the same de broglie wavelength.
determine the ratio of (i) their accelerating Potential (ii) their speed.
Sol. (i) we know λ = h/√2mqV => V = h²/2mqλ²
we have ma=4mp, qα = 2qp
∴ Vp/Va = 8/1
(ii) λ = h/mv
=> Vp/Va = ma/mp = 4:1
7. The work function for following metals is given Na = 2.75 ev,
k = 2.30 ev, Mo = 4.17ev, Ne: 5.15ev which of these metal will not emit
Photo electron for λ=3300Å from a He-Cd Laser which is 1 m away
from the metal. What happened when laser bring close to metal plate.
Sol. λ = 3300×10⁻¹⁰m
E= hc/λ = 3.75ev, so metal Na and k emit P.E. only.
(ii) If laser is bring close to plate then intensity of light increases so,
number of photoelectron ↑.
8. Ultra voilet light of λ=2271 Å° from 100W laser on M0 metal surface. If
stopping potential is 13 V. Estimate the work function. How would M0
respond to a high intensity (10⁵ Wm²) red light of wave length 6328 Å°
produced by laser.
Sol. By Einstein's law hv-Wo= K.E
hv-Wo = eVo
Wo = hv-eVo
Wo = 4.2 ev (solve yourself)
and V0 = h/λ = 1.0 x 10¹⁵ Hz
for λ = 6328 Å° = 6320 x 10⁻¹⁰ m.
∴ v = 4.74 x 10¹⁴ Hz.
Here v
9. The stopping potential is an experiment is 1.5 V. What is the maximum
K.E.
Sol. K.E = eVo
= 1.6 x 10⁻¹⁹ x 1.5
K.E = 1.5 ev.
10. If light of wave length 412.5 nm is incident on each of the metals, which
ones will show photo-electric emission and why?
Sol. E = hv/λ
= 3.01 eV
∴ Sodium (Na) and Potassium (k)
Shows photo electric effect.
11. If KE = 2 eV find stopping potential ?
Sol eVo = 2eV Vo = 2V
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