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A Proton and an Electron Are Accelerated by the Same Potential Difference. Let λE and λPdenote the De Broglie Wavelengths of the Electron and the Proton, Respectively. - Physics

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Question

A proton and an electron are accelerated by the same potential difference. Let λe and λpdenote the de Broglie wavelengths of the electron and the proton, respectively.

Options

  • λe = λp

  • λe < λp

  • λe > λp

  • The relation between λe and λp depends on the accelerating potential difference.

MCQ
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Solution

λe > λp

Let me and mp be the masses of electron and proton, respectively.
Let the applied potential difference be V.

Thus, the de-Broglie wavelength of the electron,

`λ_e = h/sqrt(2m_eeV)`    ....(1)

And de-Broglie wavelength of the proton,

`λ_p = h/sqrt(2m_peV)`        ....(2)

Dividing equation (2) by equation (1), we get :

`λ_p/λ_e = sqrt(m_e)/sqrt(m_p)`

`m_e < m_p`

`therefore λ_p/λ_e < 1`

⇒ `λ_p < λ_e`

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Chapter 20: Photoelectric Effect and Wave-Particle Duality - MCQ [Page 264]

APPEARS IN

HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 20 Photoelectric Effect and Wave-Particle Duality
MCQ | Q 14 | Page 264

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