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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

2

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B

12

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C

22

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D

None of these

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Solution

The correct option is C

22


Step 1: Given

Potential difference=V

Step 2: Determine the de-Broglie wavelength of a proton

The de-Broglie wavelength of a particle is given by λ=hP (where h is the Planck's constant and P is the momentum)

Let the mass of the proton be mp and the momentum of the proton is given by p=2mpk (where k is the kinetic energy)

The de-Broglie wavelength will be λp=h2mpk

Kinetic energy for a proton in an electric field can be rewritten as k=qpV

Therefore, λp=h2mpqpV(1)

Step 3: Determine the de-Broglie wavelength of an alpha particle

Let the mass of an alpha particle be ma

Similarly, the de-Broglie wavelength will be λa=h2maqaV(2)

Step 4: Compare the properties of a proton and an alpha-particle

qaandqp are the charges on the alpha particle and the proton.

Comparing the charges, qa=2qp

Comparing the masses, ma=4mp

Step 5: Substituting the above comparisons in equation (2)

λa=h2maqaV=h2×4ma×2qp×V(3)

Step 6: Determine the ratio of equations (1) and (3)

λpλa=h2mpqpVh2×4ma×2qp×V=42×22=22

Hence, option C is the correct option.


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