$B e^{3 +}$ and a proton are accelerated by the same potential, their de-Broglie wavelenghts have the ratio (assume mass of proton = mass of neutron):

1 : 2

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1 : 4

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1 : 1

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1 : 3 \sqrt{3}

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Solution

The correct option is C $1 : 3 \sqrt{3}$
The de-Broglie wavelength for a mass $m$ having a charge $q$ when accelerated by a potential difference $V$ is given by,
$λ_{b} = \frac{h}{\sqrt{2 m q V}}$
For the same potential,
$λ_{b} \propto \frac{1}{\sqrt{q m}}$
For proton, let $q_{1} = q_{p} = e$ and $m_{1} = m_{p} = m$
Hence, $q_{2} = q_{B e^{3 +}} = 3 e$ and $m_{2} = m_{B e^{3 +}} = 9 m$

Therefore,
$\frac{λ_{2}}{λ_{1}} = \frac{\sqrt{m e}}{\sqrt{27 m e}} = \frac{1}{3 \sqrt{3}}$

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Be3+ and a proton are accelerated by the same potential, their de-Broglie wavelenghts have the ratio (assume mass of proton = mass of neutron):

The correct option is C 1:3√3The de-Broglie wavelength for a mass m having a charge q when accelerated by a potential difference V is given by,λb=h√2mqVFor the same potential,λb∝1√qmFor proton, let q1=qp=e and m1=mp=mHence, q2=qBe3+=3e and m2=mBe3+=9mTherefore,λ2λ1=√me√27me=13√3

$B e^{3 +}$ and a proton are accelerated by the same potential, their de-Broglie wavelenghts have the ratio (assume mass of proton = mass of neutron):