Question

Question :

When a charged particle of charge x C moves through a potential Difference of y V , the gain in Kinetic energy is equal to xy J.

An electron and an alpha particle have their masses in the ratio of 1 : 7200 and charges in the ratio of 1 : 2 . If they start moving from rest through the same electrical Potential Difference , find the ratio of their Velocities.

⠀⠀⠀⠀⠀Solve it with ⠀⠀⠀Proper explanation !!


⠀⠀⠀⠀⠀⠀No spams :(​

Answer 1

Given :

If particle has charge x coulombs and potential difference y volts . Then gain in K.E. = xy joules.

Using similar analysis,

$\frac{(mass)electron}{(mass)alpha \: \: electron} = \frac{1}{7200} \\ \\ \frac{charge \: \: electron}{charge \: \: alpha \: \: electron} = \frac{1}{2}$

Since initially there at rest, initial K.E. = 0;

They move through same potential difference. Let potential difference be y volts.

So, gain in K.E = Final K.E – Initial K.E

= Final K.E – 0

So, gain in K.E = charge×potentialdifference

Ratio of gain in K.E of electron and α particle = Ratio of final K.E of electron and α particles .

$= \frac{ \frac{1}{2} {mv}^{2}_e \times y }{ \frac{1}{2} {mv}^{2} _ \alpha \times y} \\ \\ = \frac{1}{2} \times 7200 = 3600$

$\frac{V_e}{V_ \alpha } = \sqrt{3600} = 60$

•°•Ratio is 60:1

Answer 2

Given :

For an electron and an alpha particle;

Ratio of masses = 1:7200

Ratio of charges = 1:2

Both start moving from rest through the same electrical potential$.$

To Find :

We have to find ratio of their velocities.

Solution :

Mechanical kinetic energy

• K(m) = 1/2 mv²

Electrical kinetic energy

• K(e) = eV

where,

m denotes mass of particle

v denotes velocity of particle

e denotes charge of particle

V denotes potential difference

➠ $\sf{\dfrac{\frac{1}{2}\times m_1v_1^2}{\frac{1}{2}\times m_2v_2^2} = \dfrac{e_1V_1}{e_2V_2}}$

ATQ,

• m₁ : m₂ =1 : 7200
• e₁ : e₂ = 1 : 2
• V₁ = V₂ = V

➠ m₁/m₂ × v₁²/v₂² = e₁/e₂

➠ 1/7200 × v₁²/v₂² = 1/2

➠ v₁²/v₂² = 7200/2

➠ v₁/v₂ = √3600

➠ v₁ : v₂ = 60 : 1

Cheers!