# Question #1e3b6

Jun 22, 2016

Option: Deflection is proportional to charge-mass ratio at a particular instant.

#### Explanation:

Following is in support of the option selected.

(a) Let $\text{the charge of a proton } = q$
$\text{the charge of an " alpha " particle} = 2 q$

$\text{The mass of a proton } = m$

$\text{the mass of an " alpha " particle} \approx 4 m$

$E = \text{ Intensity of electric field applied}$

${v}_{p} = \text{vel. of proton after being accelerated under PD } \Delta V$

${v}_{\alpha} = \text{vel. of "alpha"-particle after being accelerated under PD } \Delta V$

So by conservation energy the particles will have KE equal to the work done on them by the potential gradient.

For proton we can write

$\frac{1}{2} m {v}_{p}^{2} = q \Delta V \implies {v}_{p} = \sqrt{\frac{2 q \Delta V}{m}}$

Similarly for $\alpha$-particle we can write

$\frac{1}{2} \cdot 4 m {v}_{\alpha}^{2} = 2 q \Delta V \implies {v}_{\alpha} = \sqrt{\frac{q \Delta V}{m}}$
We see that velocity of proton is greater than the velocity of alpha particle. If you note closely it is due to different charge to mass ratio of the particles.

As such proton will always travel faster under the given conditions.

(b) After gaining above noted velocities under the influence of same PD the particles enter in an uniform electric field (intensity E) and suffer deflections in the direction of E field as both the particles are positively charged.

Particles are projected along positive direction of x-axis and electric field E is applied towards positive direction of y-axis.

The force experienced by the proton is ${F}_{p} = q \times E$ (using scalar parts of the vector quantities)

The acceleration of proton

${a}_{\text{proton}} = {F}_{p} / m = \frac{q E}{m}$

The force experienced by the $\alpha$-particle is ${F}_{\alpha} = 2 q \times E$

Accelertion on $\alpha$ particle

${a}_{\text{alpha}} = {F}_{\alpha} / \left(4 m\right) = \frac{2 q E}{4 m} = \frac{q E}{2 m}$

So ${a}_{\text{proton">a_"alpha}}$

Here again we see that proton is accelerated faster as compared to alpha particle in the same electric field E. This leads to conclusion about difference in deflection of the two particles and its dependence on different charge to mass ratios of the particles.
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Bonus

Now lets us examine the trajectory to understand the deviations they undergo during their motions in an uniform electric field of intensity E.

The Trajectory Of Proton

Let us consider the point of projection as origin and $\left(x , y\right)$ be the position of proton after $t$ sec of its journey in the E.field.

So it suffers horizontal displacement $x \text{ in time } t$.

$\therefore x = {v}_{p} \times t = \sqrt{\frac{2 q \Delta V}{m}} \times t$

$\implies {x}^{2} = \frac{2 q \Delta V}{m} \times {t}^{2} \ldots . . \left(1\right)$

Again vertical displacement during this time will be

$y = \frac{1}{2} {a}_{p} \times {t}^{2} = \frac{1}{2} \times \frac{q E}{m} \times {t}^{2.} \ldots . . \left(2\right)$

Dividing (2) by (1) we get

$\frac{y}{x} ^ 2 = \frac{1}{4} \frac{E}{\Delta V}$

$\implies y = \frac{1}{4} \frac{E}{\Delta V} {x}^{2}$

$\implies y = k {x}^{2} , \text{where k"=1/4E/(DeltaV) " is a constant }$

The Trajectory Of $\alpha$ -particle

If $\left(x , y\right)$ be the position $\alpha$-particle after time $t$ of its journey in the E.field, taking point of projection as the origin.

So it suffers horizontal displacement $x \text{ in time } t$

$\therefore x = {v}_{\alpha} \times t = \sqrt{\frac{q \Delta V}{m}} \times t$

$\implies {x}^{2} = \frac{q \Delta V}{m} \times {t}^{2} \ldots . . \left(3\right)$

Again vertical displacement during this time will be

$y = \frac{1}{2} {a}_{\alpha} \times {t}^{2} = \frac{1}{2} \times \frac{2 q E}{4 m} \times {t}^{2.} \ldots . . \left(4\right)$

Dividing (4) by (3) we get

$\frac{y}{x} ^ 2 = \frac{1}{4} \frac{E}{\Delta V}$

$\implies y = \frac{1}{4} \frac{E}{\Delta V} {x}^{2}$

$\implies y = k {x}^{2} , \text{where k"=1/4E/(DeltaV) " is constant }$

So the two particles under consideration follow same equation of trajectory.

Quantitative explanation

By eq(2) the vertical displacement of proton in $t$ sec

${y}_{p} = \frac{1}{2} \times \frac{{q}_{p} E}{m} _ p \times {t}^{2.} \ldots \ldots . . \left(5\right)$

By eq(4) the vertical displacement of $\alpha$ - particle in $t$ sec

${y}_{\alpha} = \frac{1}{2} \times \frac{{q}_{\alpha} E}{{m}_{\alpha}} \times {t}^{2.} \ldots \ldots . \left(6\right)$

Dividing (5) by (6)

${y}_{p} / {y}_{\alpha} = \frac{{q}_{p} / {m}_{p}}{{q}_{\alpha} / {m}_{\alpha}} \ldots \ldots \left(7\right)$

So in general eq (7) reveals that the deflection of that particle is more, which has higher charge - mass ratio.

(In our case this ratio is twice for proton than for $\alpha$-particle. So proton will be more deflected.)

Jun 25, 2016

Depends on charge and mass of particle.

#### Explanation:

As both proton and alpha particle are positively charged, both will be deflected in the direction of the electric field. Let this be along the $y$-axis. Deflection of the particles will be only in the $y$ direction. Initial velocity of both are zero in this direction.

We know that when a charged particle moves in an electric field the force is given by the expression
$\vec{F} = q \vec{E}$

We need to remember that as the electric field is orthogonal, there is no change in the velocities of the particles in $x$ direction.
Using scalar parts as direction of movement is fixed.
$F = q E$
Using Newton's second law of motion
$F = m a$ and equating both we obtain
Acceleration $a = \frac{q E}{m}$

Also we know that Deflection implies change in acceleration of an object due to contact (collision) with a surface or under the influence of a field.
$\therefore$ from above equation we deduce that deflection is dependent on the charge and mass of the particle.

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The acceleration or change in acceleration translates to displacement. Therefore, deduction above can further be quantified by calculating displacement in the $y$ direction as follows.

Charge on the proton $= {q}_{p}$, and
Charge on alpha particle $= {q}_{\alpha}$

Mass of the proton $= {m}_{p}$ and
Mass of the alpha particle ${m}_{\alpha}$

Let $V$ be the accelerating potential difference before both enter the perpendicular electric field $E$.

Let both particles be at rest initially. Suppose origin coincides with the point when the particles enter the uniform electric field. Let displacement for both be zero in the $y$ direction.

Now using kinematic equation
$s = u t + \frac{1}{2} a {t}^{2}$, we obtain the deflection in $y$ direction as
$y = \frac{1}{2} \frac{q E}{m} {t}^{2}$
For proton
${y}_{p} = \frac{1}{2} \frac{{q}_{p} E}{m} _ p {t}^{2}$ ......(1)
and for alpha particle
${y}_{\alpha} = \frac{1}{2} \frac{{q}_{\alpha} E}{m} _ \alpha {t}^{2}$ .......(2)

Dividing (1) by (2) we obtain
${y}_{p} / {y}_{\alpha} = \frac{\frac{1}{2} \frac{{q}_{p} E}{m} _ p {t}^{2}}{\frac{1}{2} \frac{{q}_{\alpha} E}{m} _ \alpha {t}^{2}}$
${y}_{p} = {q}_{p} / {m}_{p} / {q}_{\alpha} / {m}_{\alpha} \times {y}_{\alpha}$ ......(3)

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Charge on the proton $= {e}^{+}$, and
Charge on alpha particle $= 2 {e}^{+}$

For sake of simplicity lets us take ${m}_{\alpha} \approx 4 {m}_{p}$
Therefore equation (3) becomes
${y}_{p} = \frac{e \times 4 {m}_{p}}{2 e \times {m}_{p}} \times {y}_{\alpha}$
${y}_{p} = 2 {y}_{\alpha}$

We see that for the same time of travel, displacement of proton is twice the displacement of alpha particle which is approximately equal to $e / m$ ratio of the two.

Also from above we notice that displacement for proton is more. However, even though applicable for proton-alpha particle example, this is not the correct answer option.