Question

# A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is:

A

$\sqrt{2}$

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B

$\frac{1}{\sqrt{2}}$

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C

$2\sqrt{2}$

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D

None of these

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Solution

## The correct option is C $2\sqrt{2}$Step 1: GivenPotential difference=$V$Step 2: Determine the de-Broglie wavelength of a protonThe de-Broglie wavelength of a particle is given by $\lambda =\frac{h}{P}$ (where $h$ is the Planck's constant and $P$ is the momentum)Let the mass of the proton be ${m}_{p}$ and the momentum of the proton is given by $p=\sqrt{2{m}_{p}k}$ (where $k$ is the kinetic energy)The de-Broglie wavelength will be ${\lambda }_{p}=\frac{h}{\sqrt{2{m}_{p}k}}$Kinetic energy for a proton in an electric field can be rewritten as $k={q}_{p}V$Therefore, ${\lambda }_{p}=\frac{h}{\sqrt{2{m}_{p}{q}_{p}V}}\left(1\right)$Step 3: Determine the de-Broglie wavelength of an alpha particleLet the mass of an alpha particle be ${m}_{a}$Similarly, the de-Broglie wavelength will be ${\lambda }_{a}=\frac{h}{\sqrt{2{m}_{a}{q}_{a}V}}\left(2\right)$Step 4: Compare the properties of a proton and an alpha-particle${q}_{a}and{q}_{p}$ are the charges on the alpha particle and the proton.Comparing the charges, ${q}_{a}=2{q}_{p}$Comparing the masses, ${m}_{a}=4{m}_{p}$Step 5: Substituting the above comparisons in equation $\left(2\right)$${\lambda }_{a}=\frac{h}{\sqrt{2{m}_{a}{q}_{a}V}}\phantom{\rule{0ex}{0ex}}=\frac{h}{\sqrt{2×4{m}_{a}×2{q}_{p}×V}}\left(3\right)$Step 6: Determine the ratio of equations $\left(1\right)$ and $\left(3\right)$$\frac{{\lambda }_{p}}{{\lambda }_{a}}=\frac{\frac{h}{\sqrt{2{m}_{p}{q}_{p}V}}}{\frac{h}{\sqrt{2×4{m}_{a}×2{q}_{p}×V}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}$Hence, option C is the correct option.

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