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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.
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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