Question

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

पर्याय

• λ_e = λ_p

• λ_e < λ_p

• λ_e > λ_p

• The relation between λ_e and λ_p depends on the accelerating potential difference.

Answer 1

λ_e > λ_p

Let m_e and m_p 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`