Question

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A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will be

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Answer 1

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$v=\sqrt{\frac{ev}{m} }$

The helium ion has:

a potential difference in vacuum = V,

Charge = 2e

and mass = 4m,

speed = v,

mass = 4m

From electrostatics, the work done is the product of charge and its potential difference.

Therefore, Work done = charge × potential difference = 2e × V = 2eV

This work done is in form of kinetic energy, therefore:

Kinetic energy = 1/2 × mass × speed²

⇒  Work done = Kinetic energy

$2eV=\frac{1}{2} *4m*v^2=2m*v^2\\v^2=\frac{2eV}{2m}\\ v^2=\frac{eV}{m}\\ v=\sqrt{\frac{ev}{m} }$

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Answer 2

The final speed of the ion will be √(eV/m).

Given,

A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum.

To Find,

The final speed of the ion.

Solution,

Since we know that the work done is the product of charge and its potential difference.

So,

Work done = charge × potential difference = 2e × V = 2eV

This work done is in form of kinetic energy, therefore

Work done = Kinetic energy

Substituting the values

2ev = 1/2 mv²

2ev = 1/2 4mv²

v² = eV/m

v = √(eV/m)

Hence, the final speed of the ion will be √(eV/m).

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