Question

Tell me please today is my exam!

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

$\red{\bold{{\underline{ Answer \: with \: Explanation \: :}}}}$

Keep this formula in Mind :-

$\huge\pink{\bold{v=\sqrt{\frac{ev}{m}}}}$

Given in the Question :-

• Potential difference in vacuum = V
• Charge = 2e
• 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 =

$\green{\bold {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:2e × V = 2eV}}$

This Work Done is in form of Kinetic Energy.

Therefore :-

Kinetic energy = 1/2 × mass × speed²

⇒  Work done = Kinetic energy

$\to \bf{ \red{2eV = \dfrac{1}{2} \times 4m \: v^{2} = 2m \times v ^{2} }} \\ \\ \to \bf{ \red{v ^{2} = \dfrac{eV}{m} }} \\ \\ \to \bf{ \red{v = \sqrt{ \dfrac{eV}{m} } }}$

Answer 2

The final speed will be $\sqrt{\frac{eV}{m} } .$

Given:

Mass of helium ion= 4m

Charge= 2e

Potential difference= V

To Find:

The final speed of the helium ion

Solution:

We know that,

Work done is the product of the potential difference and the charge of an ion.

Therefore, work done= V x 2e= 2eV

This work can be expressed in the form of kinetic energy.

So, kinetic energy= work done= 2eV

Also,

Kinetic energy= (1/2) x (mass) x (speed)²

or, 2eV= (1/2) x 4m x speed²

or, $speed^{2} = \frac{2eVX2}{4m}$

or, $speed= \sqrt{\frac{eV}{m} }$

Hence, the final speed will be      $\sqrt{\frac{eV}{m} } .$

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