Question

An electron and a proton are possessing the same amount of kinetic energy. Which of the two have greater wavelength?

Answer 1

We are actually considering that the particles are non-relativistic.

The mass of a proton is approximately about 1800 times the mass of an electron

Given that, the kinetic energies of the electron and proton are equal, we can use the formula  

$KE\quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }$

${ KE }_{ e }\quad =\quad { KE }_{ p }$

$\frac { 1 }{ 2 } { m }_{ e }{ v }_{ e }^{ 2 }\quad =\quad \frac { 1 }{ 2 } 1800{ m }_{ e }{ v }_{ p }^{ 2 }$

On simplification we get

${ v }_{ e }^{ 2 }\quad =\quad 1800{ v }_{ p }^{ 2 }$

Taking square root on both sides, we get

${ v }_{ e }\quad \sim \quad 42{ v }_{ p }$

Formula for momentum, (p) = mv

So,

${ p }_{ e }\quad =\quad { m }_{ e }{ v }_{ e }$

${ p }_{ p }\quad =\quad { m }_{ p }{ v }_{ p }\quad \sim \quad 1800{ m }_{ e }\quad \times \quad \frac { { v }_{ e } }{ 42 } \quad =\quad 42{ m }_{ e }{ v }_{ e }$

So, ${ p }_{ p }\quad =\quad 42{ p }_{ e }\quad .....(1)$

The "de Broglie wavelength" of a particle is "dependent" on its "momentum".  

$De\quad Broglie\quad wavelength\quad \alpha \quad \frac { 1 }{ p } \quad .....(2)$

From (1) and (2), we can conclude that the electron's wavelength is around 42 times the proton's wavelength and hence, the electron's wavelength is longer than the proton's wavelength.