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Question: An electron of mass m when accelerated through a potential difference V, has de-Broglie wavelength l...

An electron of mass m when accelerated through a potential difference V, has de-Broglie wavelength l. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be –

A

λmM\frac{\lambda m}{M}

B

lmM\sqrt{\frac{m}{M}}

C

λMm\frac{\lambda M}{m}

D

lMm\sqrt{\frac{M}{m}}

Answer

lmM\sqrt{\frac{m}{M}}

Explanation

Solution

Momentum of electron pe =2meV\sqrt{2meV}

Momentum of proton pp =2MeV\sqrt{2MeV}

\ λpλe\frac{\lambda_{p}}{\lambda_{e}}= h/pph/pe\frac{h/p_{p}}{h/p_{e}} = pepp\frac{p_{e}}{p_{p}} = 2meV2MeV\frac{\sqrt{2meV}}{\sqrt{2MeV}}= mM\sqrt{\frac{m}{M}}

\ lp = lemM\sqrt{\frac{m}{M}}= lmM\sqrt{\frac{m}{M}}