$B e^{3 +}$ and a proton are accelerated by the same potential, their de – Broglie wavelengths have the ratio (assume mass of proton = mass of neutron) $x : y$ . Then $x + y$ is:

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Solution

Mass of the proton = 1 amu
mass of $B e^{3 +} = 9$ amu
From de Broglie equation,
$λ = \frac{h}{\sqrt{2 m E}}$
$λ \propto \frac{1}{\sqrt{m}}$
Hence
$\frac{λ_{B e^{3 +}}}{λ_{H}} = \frac{\sqrt{1}}{\sqrt{9}} = 1 : 3$
Hence $x + y = 4$

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Be3+ and a proton are accelerated by the same potential, their de – Broglie wavelengths have the ratio (assume mass of proton = mass of neutron) x:y. Then x+y is:

Mass of the proton = 1 amu mass of Be3+=9 amu From de Broglie equation, λ=h√2mE λ∝1√m Hence λBe3+λH=√1√9=1:3 Hence x+y=4

$B e^{3 +}$ and a proton are accelerated by the same potential, their de – Broglie wavelengths have the ratio (assume mass of proton = mass of neutron) $x : y$ . Then $x + y$ is:

Mass of the proton = 1 amu
mass of $B e^{3 +} = 9$ amu
From de Broglie equation,
$λ = \frac{h}{\sqrt{2 m E}}$
$λ \propto \frac{1}{\sqrt{m}}$
Hence
$\frac{λ_{B e^{3 +}}}{λ_{H}} = \frac{\sqrt{1}}{\sqrt{9}} = 1 : 3$
Hence $x + y = 4$