An \alpha-particle (charge +2e) and a \beta^--particle (charge -e) emitted from a natural … — JAMB UTME

An $\alpha$ -particle (charge $+2e$ ) and a $\beta^-$ -particle (charge $-e$ ) emitted from a natural radioactive source enter a region of uniform electric field $E$ perpendicularly with equal initial kinetic energies. If $y_\alpha$ and $y_\beta$ represent the magnitudes of their transverse deflections after traveling the same horizontal distance through the field, what is the ratio $\frac{y_\beta}{y_\alpha}$ ?

$\frac{1}{2}$ Cevap
B
$2$
C
$\frac{m_\alpha}{2 m_\beta}$
D
$1$

Cevap

The ratio of the transverse deflections

\frac{y_\beta}{y_\alpha}

\frac{1}{2}

For a charged particle entering a uniform electric field perpendicularly with kinetic energy

K

, the transverse deflection is

y = \frac{|q| E L^2}{4K}

. Because both particles enter with identical kinetic energies and travel the same horizontal distance

L

, the deflection is directly proportional to the magnitude of charge

|q|

and independent of mass. Since the

\beta^-

-particle has charge magnitude

e

and the

\alpha

-particle has charge magnitude

2e

, the ratio of their deflections is

\frac{e}{2e} = \frac{1}{2}

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Express initial horizontal velocity in terms of kinetic energy

K

and mass

m

v_x = \sqrt{\frac{2K}{m}}

, so time spent in the electric field of horizontal length

L

t = \frac{L}{v_x} = L \sqrt{\frac{m}{2K}}

Both particles enter horizontally with equal initial kinetic energy

K

Derive the formula for transverse deflection

y

in a uniform electric field

E

The transverse force is

F_y = |q|E

, giving acceleration

a_y = \frac{|q|E}{m}

. The deflection is

y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left(\frac{|q|E}{m}\right) \left(\frac{L^2 m}{2K}\right) = \frac{|q|E L^2}{4K}

The particle mass

m

cancels out when time is expressed in terms of kinetic energy.

Calculate

y_\alpha

and

y_\beta

using their respective charge magnitudes

|q_\alpha| = 2e

and

|q_\beta| = e

y_\alpha = \frac{2e E L^2}{4K} = \frac{e E L^2}{2K}

and

y_\beta = \frac{e E L^2}{4K}

\alpha

-particle carries a charge of

+2e

while a

\beta^-

-particle carries a charge of

-e

Compute the ratio

\frac{y_\beta}{y_\alpha}

\frac{y_\beta}{y_\alpha} = \frac{\frac{e E L^2}{4K}}{\frac{e E L^2}{2K}} = \frac{1}{2}

Dividing

y_\beta

y_\alpha

cancels all terms except the ratio of charge magnitudes.

Anahtar Kavram

Deflection of charged radioactive emissions in a uniform electric field under equal kinetic energy

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