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Dual Nature of Radiation and Matter

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Physics Part II

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Class 10 Class 12

When a metallic surface is illuminated with radiation of wavelength straight lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 straight lambda, the stopping potential is V/4. The threshold wavelength for the metallic surface is,

  • 5 straight lambda
  • 5 over 2 lambda
  • 3straight lambda

  • 4straight lambda


C.

3straight lambda

When a metallic surface is illuminated with radiation of wavelength straight lambda, the stopping potential is V.

Photoelectric equation can be written as,

eV equals space space hc over straight lambda minus hc over straight lambda subscript straight o        ... (i)
Now, when the same surface is illuminated with radiation of wavelength 2straight lambda, the stopping potential is V/4. So, photoelectric equation can be written as,

space space space space space space eV over 4 equals fraction numerator h c over denominator 2 lambda end fraction minus fraction numerator h c over denominator lambda subscript o end fraction
rightwards double arrow space e V space equals space fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator italic 4 h c over denominator lambda subscript o end fraction space space space space... space left parenthesis i i right parenthesis
From equations (i) and (ii), we get

space space space hc over straight lambda minus hc over straight lambda subscript straight o equals fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator 4 h c over denominator straight lambda subscript straight o end fraction
rightwards double arrow space 1 over straight lambda minus 1 over straight lambda subscript straight o equals italic 2 over lambda minus 4 over straight lambda subscript straight o
rightwards double arrow space space straight lambda subscript straight o space equals space 3 straight lambda

When a metallic surface is illuminated with radiation of wavelength straight lambda, the stopping potential is V.

Photoelectric equation can be written as,

eV equals space space hc over straight lambda minus hc over straight lambda subscript straight o        ... (i)
Now, when the same surface is illuminated with radiation of wavelength 2straight lambda, the stopping potential is V/4. So, photoelectric equation can be written as,

space space space space space space eV over 4 equals fraction numerator h c over denominator 2 lambda end fraction minus fraction numerator h c over denominator lambda subscript o end fraction
rightwards double arrow space e V space equals space fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator italic 4 h c over denominator lambda subscript o end fraction space space space space... space left parenthesis i i right parenthesis
From equations (i) and (ii), we get

space space space hc over straight lambda minus hc over straight lambda subscript straight o equals fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator 4 h c over denominator straight lambda subscript straight o end fraction
rightwards double arrow space 1 over straight lambda minus 1 over straight lambda subscript straight o equals italic 2 over lambda minus 4 over straight lambda subscript straight o
rightwards double arrow space space straight lambda subscript straight o space equals space 3 straight lambda

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What is the de-Broglie wavelength of
(a)    a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s.
(b)    a ball of mass 0.060 kg moving at a speed of 1.0 m/s and
(c)    a dust particle of mass 1.0 x 10–9 kg drifting with a speed of 2.2 m/s?


(a) Given,

Mass of the bullet, m = 0.040 kgVelocity with which the bullet is travelling, v = 1 kms-1 = 103 ms-1Using the formula of momentum,                                                               p = mv                                                                  = 0.040 × 103                                                                   = 40 kg ms-1
 De-broglie wavelength is, 
                         λ = hp     = 6.62 × 10-3440      = 1.7 × 10-35m 


(b) Mass of the ball, m = 0.060 kg 
     Velocity with which the ball is moving, v = 1.0 ms-1
     Momentum of the particle,  p = mv = 0.060 kg ms-1

Therefore, 

De-broglie wavelength of the particle, λ = hp    = 6.62 × 10-340.060    = 1.1 × 10-32

(c) Mass of the dust particle, m = 1.0 × 10-9 kg
     Velocity of the particle, v= 2.2 m/s Momentum of the particle, p = mv = 2.2 × 10-9Therefore,De-broglie wavelength, λ = hp = 6.62 × 10-342.2 × 10-19                                                             = 3 × 10-25m.

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Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
a.) Find the energy and momentum of each photon in the light beam. 

b.) How many photons per second, on the average arrive at a target irradiated by this beam?

c.) How fast does a hydrogen atom have to travel in order to have the same  momentum as that of the photon?

Given,
Wavelength of monochromatic light, λ = 632.8 nm = 632.8 x 10
–9 m
 Frequency, v = cλ = 3 × 108632.8 × 10-9Hz 

                                 = 4.74 × 1014Hz

(a) Energy of a photon, E = hv

                                       = 6.63 × 10-34× 4.74 × 1014J= 3.14 × 10-19J.

Momentum of each photon, p (momentum) = hλ                            = 6.63 × 10-34632.8 × 10-9                              = 1.05 × 10-27 kg ms-1 

(b) Power emitted, P = 9.42 mW = 9.42 x 10–3 W
Now, P = nE  

This implies, 

n = PE = 9.42 × 10-3W3.14 × 10-19J                = 3 × 1016 photons/sec. 
Thus, these many number of protons arrive at the target.

(c) Velocity of hydrogen atom 

                    = Momentum 'p' of H2 atom (mv)Mass of H2 atom(m) 

                  v = 1.05 × 10-271.673 × 10-27ms-1   

                    = 0.63 ms-1. 
Thus, the hydrogen atom travel at a speed of 0.63 m/s  to have the same  momentum as that of the photon.
360 Views

The energy flux of sunlight reaching the surface of the earth is 1.388 x 103 W/m2. How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Energy flux of sunlight = Total energy per square metre per second = 1.388 x 103 Wm–2 

Energy of each photon is given by,

                       E = hcλ 
Therefore, 
 Power of each photon, P  = nE                                               = 6.63 × 10-34× 3 × 108550 × 10-9J                                              = 3.62 × 10-19J 
Number of photons incident on earth's surface per square metre per second is, 

= Total energy per square metre per secondEnergy of one photon 

= 1.388 × 1033.62 × 10-19 = 3.8 × 1021.

198 Views

Calculate the
(a)    momentum, and
(b)    de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V. 


Given, 
Potential difference, V = 56 V

Energy of electron accelerated, 
= 56 eV = 56 × 1.6 × 10-19J
(a) As, Energy, E = p22m               [p = mv,  E = 12mv2]
                  p2 = 2mE 

                   p = 2mE 

                   p = 2 × 9 × 10-31 × 56 × 1.6 × 10-19 

                       p = 4.02 × 10-24 kg ms-1 
is the momentum of the electron. 

(b) Now, using De-broglie formula we have,  p = hλ
        λ = hp = 6.62 × 10-344.02 × 10-24 

              = 1.64 × 10-10m = 0.164 × 10-9m 
 
i.e.,      λ = 0.164 nm , is the De-broglie wavelength of the electron.
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What is the
(a)    momentum,
(b)    speed, and
(c)    de-Broglie wavelength of an electron with kinetic energy of 120 eV.


given, 
Kinetic energy of the electron, K.E = 120 eV

(a) Momentum of the electron is given as, p  = 2mE 

 p = 2 ×(9 × 10-31) × (120 × 1.6 × 10-19)    = 5.88 × 10-24 kg ms-1

(b) Now, since p = mv 
we have,      
                     v = pm    = 5.88 × 10-249.1 × 10-31 
                              
                 v = 6.46 × 106 m/s 

(c) De- broglie wavelength of electron is given by, 
                 λ = hp    = 6.63 × 10-345.88 × 10-24    = 1.13 × 10-10m

                    = 1.13 Å.                



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