$B e^{+ 3}$ & a proton are accelerated same potential their de-Broglie wavelength have the ratio:

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Solution

$m =$ mass

$q =$ charge

$V =$ potential

$λ = h / m v$

$λ_{p r o t o n} = h / m v$

The momentum acquired by the proton = $(2 m q V)^{1 / 2}$

Thus $λ_{p r o t o n} = h / (2 m q V)^{1 / 2}$

For $λ_{B e^{+ 3}} = h / (2 \times 9 m \times 3 q \times V)^{1 / 2}$

Then $λ_{B e^{+ 3}} / λ_{p r o t o n} = 1 / 3 (3)^{1 / 2}$

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Be+3 & a proton are accelerated same potential their de-Broglie wavelength have the ratio:

m= massq= charge V= potentialλ=h/mvλproton=h/mvThe momentum acquired by the proton = (2mqV)1/2Thus λproton=h/(2mqV)1/2For λBe+3=h/(2×9m×3q×V)1/2Then λBe+3/λproton=1/3(3)1/2

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