An alpha particle is accelerated through a potential difference of v volts from rest. The de brogil wavelength associated with it is?

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Answer

Calculation of de Broglie wavelength of alpha particle accelerated through a potential difference of v volts

The de Broglie wavelength (λ) associated with a particle is given by the equation:

λ = h/p

where h is Planck's constant and p is the momentum of the particle. For an alpha particle, the momentum can be calculated as:

p = mv

where m is the mass of the alpha particle and v is the velocity it acquires after being accelerated through a potential difference of v volts. The velocity can be calculated using the equation for the kinetic energy:

KE = ½ mv² = qv

where q is the charge on the alpha particle. Since an alpha particle has a charge of +2e, where e is the elementary charge, the kinetic energy can be expressed as:

KE = (2e)(v)

Setting this equal to the potential energy gained by the alpha particle, we have:

qV = (2e)(v)

or

v = (qV)/(2m)

Substituting this expression for v in the equation for the momentum, we get:

p = (2mqV)/(2m) = qV

Finally, substituting the expression for p in the equation for the de Broglie wavelength, we get:

λ = h/p = h/(qV)

Explanation

The de Broglie wavelength of a particle is inversely proportional to its momentum. Therefore, the higher the momentum of a particle, the smaller its de Broglie wavelength. In the case of an alpha particle accelerated through a potential difference, the momentum of the particle can be calculated using the equations for kinetic and potential energy. Substituting this expression for momentum in the equation for the de Broglie wavelength, we can calculate the wavelength associated with the alpha particle. This calculation is useful in understanding the wave-particle duality of matter and its behavior in different physical systems.

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