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Karnataka Board PUCPUC Science 2nd PUC Class 12

A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?

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Question

A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp and λa related to each other?

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Solution

In this problem since both proton and α-particle are accelerated through the same potential difference,

We know that, `λ = h/sqrt(2mqv)`

∴ `λ  oo 1/sqrt(mq)`

`λ_p/λ_α = sqrt(m_αq_α)/(m_pq_P) = sqrt(4m_p xx 2e)/sqrt(m_p xx e) = sqrt(8)`

∴ `λ_p = sqrt(8)λ_α`

i.e., wavelength of proton is `sqrt(8)` times wavelength of α-particle.

Important point: De-Broglie wavelength associated with the charged particles: The energy of a charged particle accelerated through potential difference V is `E = 1/2 mv^2 = qV`

Hence de-Broglie  wavelength `λ = h/p = h/sqrt(2mE) = h/sqrt(2mqV)`

`λ_("Electron") = 12.27/sqrt(V) Å, λ_("Proton") = 0.286/sqrt(V) Å`

`λ_("Deutron") = 0.202/sqrt(V) Å, λ_(a"-particle") = 0.101/sqrt(V) Å`

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Chapter 11: Dual Nature Of Radiation And Matter - Exercises [Page 71]

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NCERT Exemplar Physics Exemplar [English] Class 12
Chapter 11 Dual Nature Of Radiation And Matter
Exercises | Q 11.14 | Page 71

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