Question

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.

Answer 1

Wavelength of electron is greater than the wavelength of proton

Explanation:

The de Broglie wavelength $\lambda = \frac{Planck's constant }{Momentum}$

$\lambda = \frac{h}{p} = \frac{h}{mv}$

Where, $m = \textrm{mass of the particle}$

$v = \textrm{velocity of the particle}$

Now, we can say that $\lambda \propto \frac{1}{m}$

so, $\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e}$  

Where, Mass of electron $m_e = 9.1\times10^{-31} kg$

mass of proton $m_p = 1.6\times10^{-21} kg$

$\frac{\lambda_e}{\lambda_p} = \frac{1.6 \times 10^{-21}}{9.1 \times 10^{-31}}$

$\lambda_e = 1.75 \times10^{9} \lambda_p$

Hence, the wavelength of electron is greater than the wavelength of proton

Answer 2

Answer is Option (c)

Explanation:

The de Broglie wavelength is given by

$\lambda = \frac {h}{p} = \frac {h}{mv}$ """(1)

Where h = Planck’s constant; m = mass of the particle; p = momentum and v = velocity of the particle

Mass of electron, $m _e = 9.1 \times 10^-^3^1 kg$; mass of proton, $m_p = 1.6 \times 10^-^2^1 kg$

So, $\lambda \propto \frac {1}{m}$

or $m \propto \frac {1}{\lambda}$"""(2)

Hence, $\frac {1}{\lambda_e} < \frac {1}{\lambda_p} (As\ m_e < m_p)$

Finally,  

$\lambda_p} < \lambda_e$

Therefore, the wavelength of electron is greater than the wavelength of proton