An \alpha-particle (charge +2e, mass m_\alpha) and a \beta^--particle (charge -e, mass m_\… — JAMB UTME

An $\alpha$ -particle (charge $+2e$ , mass $m_\alpha$ ) and a $\beta^-$ -particle (charge $-e$ , mass $m_\beta$ ) emitted during natural radioactive decay enter a region containing uniform, mutually perpendicular electric ( $E$ ) and magnetic ( $B$ ) fields. Both particles move along paths perpendicular to both fields and traverse the region without undergoing any deflection. What is the ratio of the kinetic energy of the $\alpha$ -particle to that of the $\beta^-$ -particle, $\frac{E_{k,\alpha}}{E_{k,\beta}}$ ?

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

Answer

The ratio of their kinetic energies is equal to the ratio of their masses,

\frac{m_\alpha}{m_\beta}

In crossed uniform electric and magnetic fields acting as a velocity selector, a charged particle passes undeflected when the electric force

qE

balances the magnetic Lorentz force

qvB

. Equating these forces yields

v = \frac{E}{B}

, which depends only on field strengths and is independent of mass and charge. Because both the

\alpha

-particle and

\beta^-

-particle traverse undeflected, both possess the same speed

v

. Substituting equal speeds into the kinetic energy formula

E_k = \frac{1}{2}mv^2

leaves the ratio of their kinetic energies equal to the ratio of their rest masses,

\frac{m_\alpha}{m_\beta}

Step-by-Step Solution

Apply the force balance condition for particles moving undeflected in crossed electric and magnetic fields.

Electric force magnitude

F_E = qE

must equal magnetic force magnitude

F_B = qvB

, giving

qE = qvB \implies v = \frac{E}{B}

For zero net deflection, the electrostatic force and magnetic Lorentz force must be equal in magnitude and opposite in direction.

Determine the velocities of the

\alpha

-particle and

\beta^-

-particle.

v_\alpha = \frac{E}{B}

and

v_\beta = \frac{E}{B}

, so

v_\alpha = v_\beta = v

The velocity selection equation

v = \frac{E}{B}

is completely independent of particle mass

m

and charge

q

Express the ratio of the kinetic energies of the two emissions using

E_k = \frac{1}{2}mv^2

\frac{E_{k,\alpha}}{E_{k,\beta}} = \frac{\frac{1}{2} m_\alpha v^2}{\frac{1}{2} m_\beta v^2} = \frac{m_\alpha}{m_\beta}

Since the speed

v

is identical for both particles, the

\frac{1}{2}v^2

terms cancel out completely.

Key Concept

Velocity selector behavior and kinetic energy dependence of radiation emissions in electromagnetic fields

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