Question

Deduce the condition when the de broglie wavelength associated with an electron would be equal to that associated with a proton if a proton is 1836 times heavier than an electeon

Answer 1

Answer:

when velocity of electron is equal to  1836 times the velocity of proton

Explanation:

Data:

mass of electron = Me

mass of proton  =  Mp

velocity of electron = Ve

velocity of proton = Vp

Mass of proton = 1836 × mass of electron

Required:

The condition at which λe=λp

Calculation:

We know that

de-Broglie wavelength for electron (λe) = h / ( Me × Ve )

Similarly

de-Broglie wavelength for proton (λp) = h / ( Mp × Vp )

                                                                  =h / 1836 × Me × Vp

∵ (Mp = 1836 × Me)

The condition when the de-Broglie wavelength associated with an electron would be equal to that of proton  that is

                                           λe = λp

Putting values from above equations we get

             h / ( Me × Ve ) =h / 1836 × Me × Vp

Dividing by h and multiplying by Me on both sides we get

                         1 /   Ve =1 / ( 1836 Vp )

⇒                                Ve = 1836 Vp

Hence when velocity of electron is equal to  1836 times the velocity of proton then

    ''  The de-Broglie wavelength associated with an electron would be equal to that of proton that is

                                           λe = λp                ''

Answer 2

Answer is Ve = 1836 Vp

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