A proton and an electron are accelerated by the same potential difference. Let λ_e and λ_p denote the de Broglie wavelengths of the electron and the proton, respectively.
(a) λ_e = λ_p
(b) λ_e < λ_p
(c) λ_e > λ_p
(d) The relation between λ_e and λ_p depends on the accelerating potential difference.

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Solution

(c) 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,
$λ_{e} = \frac{h}{\sqrt{2 m_{e} e V}} . . . (1)$
And de-Broglie wavelength of the proton,
$λ_{p} = \frac{h}{\sqrt{2 m_{p} e V}} . . . (2)$
Dividing equation (2) by equation (1), we get:
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}$
m_e < m_p
$∴$ $\frac{λ_{p}}{λ_{e}} < 1$
$\Rightarrow λ_{p} < λ_{e}$

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A proton and an electron are accelerated by the same potential difference. Let λe and λp denote the de Broglie wavelengths of the electron and the proton, respectively. (a) λe = λp (b) λe < λp (c) λe > λp (d) The relation between λe and λp depends on the accelerating potential difference.