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De Broglie's Hypothesis

De-Broglie wa...

Question

De-Broglie wavelength of two particles $A$ and $B$ are plotted against $(\frac{1}{\sqrt{V}})$ ; where $V$ is potential on the particle. Which of the following relation is correct about mass of particle.

A

M_{A} = M_{B}

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B

M_{A} > M_{B}

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C

M_{A} < M_{B}

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D

M_{A} \leq M_{B}

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Solution

The correct option is C $M_{A} > M_{B}$
K.E = eV. (where v is potential)
$P = \sqrt{2 m k} = \sqrt{2 m e V}$
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m_{e} V}}$
$λ = (\frac{1}{\sqrt{v}}) . \frac{h}{\sqrt{2 m_{e}}}$
So according to above equation slope of equation is inversely proportional to mass of particle
$∵ s l o p e_{A} < s l o p e_{B}$
$∴ M_{A} > M_{B}$

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