Question

An electron and proton are acceleration through same potential difference. Which one of them has (i) less momentum (ii) Greater wave length .

Answer 1

Due to Potential difference charge will accelerate and acquire kinetic energy

$eV=\frac{1}{2}m{v}^2$

In term of momentum, $P=mv=\sqrt{2m\left(eV\right)}$

Ratio of moment of proton and electron, $\frac{P_P}{P_e}=\frac{\sqrt{2m_p\left(eV\right)}}{\sqrt{2m_e\left(eV\right)}}=\frac{\sqrt{m_p}}{\sqrt{m_e}}=42.83$

$P_p=42.83P_e$ (Momentum of proton is 42.83 times larger than momentum of electron)

From De-Broglie’s relation, $\lambda =\frac{h}{P}=\frac{h}{mv}$

Wavelength ratio, $\frac{{\lambda }_p}{{\lambda }_e}=\frac{h/P_P}{h/P_e}=\frac{P_e}{p_p}=\frac{1}{42.83}$

${\lambda }_e=42.83{\lambda }_p$ (Wavelength of electron is 42.83 times the wavelength of proton)

Hence, electron momentum is less and its wavelength is large compare to proton