Question

An electron & a proton have their masses in the ratio 1:1840. What will be the ratio of their kinetic energy if they have equal momentum ?

Answer 1

$E = \frac{1}{2} mv {}^{2}$

$p = mv = \sqrt{2mE} = \sqrt{2mqv}$

De Broglie wavelength associated with change particle

$λ = \frac{h}{p} = \frac{h}{\sqrt{2mE} } = \frac{h}{ \sqrt{mqv} }$

$E = \frac{1}{2} mv {}^{2} = K.E.P \: is \: the \: momentum \: of \: change$

$So, λP= \frac{h}{\sqrt{2mpE} }$

$So, λo= \frac{h}{\sqrt{2moE} }$

$So,  \frac{λp}{λo} = \frac{ \sqrt{me} }{ \sqrt{mp} }$

$\sqrt{ \frac{me}{mp} }$

$= \sqrt{ \frac{9.1 \times 10 {}^{ - 13} }{1.67 \times 10 {}^{ - 27} } }$

$\sqrt{ =5.449×10 {}^{−4} } \\ </p><p></p><p></p><p>$

$2.33 \times 10 {}^{ - 2}$

$=0.0233$

$= \frac{1}{43}$

And the answer is 1:43

Answer 2

Mass of electron/ Mass of proton=1:1840

K.E. = p²/2m

K.E. of electron/ K.E. of proton= p²/2m₁ × p²/2m₂

p² cancels out as momentum of electron and proton are equal

K.E. of electron/ K.E. of proton= m₂/ m₁

K.E. of electron/ K.E. of proton= 1840/1

K.E. of electron: K.E. of proton= 1840:1