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Wave Particle Duality: Matter Waves

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Question
ICSEENIPH12029761
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What is the relation between wavelength and momentum of moving particles?

Solution

straight p space equals space straight h over straight lambda ; 
where, 
p = momentum 
h = planck's constant 
straight lambda = wavelength

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Question
ICSEENIPH12029837
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Calculate the energy released when an electron annihilates a positron.

Solution

Energy of equivalent of mass of electron and positron.

Eo = moC2 
    = (9.1 x 10-31 X3 x 108 )2 
   
    = 8.2 space straight x space 10 to the power of negative 14 end exponent space equals space fraction numerator 8.2 space straight x space 10 to the power of negative 14 end exponent over denominator 1.6 space straight x space 10 to the power of negative 13 end exponent end fraction space equals space 0.51 space MeV
Energy released = 0.51 x 2 = 1.02 MeV

Question
ICSEENIPH12029845
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Which one of the following graphs in Figure 1 represents variation of de Broglie wavelength (λ) of particle having linear momentum p: 

Solution

D.

Question
ICSEENIPH12029903
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De Broglie wavelength of electrons of kinetic energy E is λ. What will be its value if kinetic energy of electrons is made 4E?

Solution

De Broglie’s wavelength of electron = λ

Let the new wavelength be λ2

K.E. of electrons = E

straight lambda over straight lambda subscript 2 space equals space fraction numerator square root of E over denominator square root of 4 E end root end fraction s i n c e space lambda space equals space fraction numerator h over denominator square root of 2 m E end root end fraction straight lambda over straight lambda subscript 2 space equals space 1 half space straight lambda subscript 2 space equals space 2 space straight lambda

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