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Step-by-Step Solution
Step 1: State the de Broglie Relation
The de Broglie wavelength λ associated with any particle is given by:
$ \lambda = \frac{h}{p} $
where $h$ is Planck’s constant, and $p$ is the momentum of the particle.
Step 2: Relate Momentum to Mass and Velocity
The momentum $p$ of a particle of mass $m$ moving with velocity $v$ is:
$ p = mv \,.
$
Thus, the de Broglie wavelength can be written as:
$ \lambda = \frac{h}{mv} \,.
$
Step 3: Use the Circular Motion Condition in a Magnetic Field
An α-particle (charge $q = 2e = 2 \times 1.6 \times 10^{-19}\,\text{C}$) moving in a uniform magnetic field $B$ follows a circular path if the magnetic force provides the centripetal force. Hence,
$ \frac{mv^2}{r} = q v B \,.
$
Rearranging gives:
$ mv = qBr \,.
$
Step 4: Substitute Known Values to Find $mv$
Given:
Radius, $r = 0.83 \times 10^{-2}\,\text{m}$
Magnetic field, $B = 0.25\,\text{Wb/m}^2$
Charge of α-particle, $q = 2 \times 1.6 \times 10^{-19}\,\text{C}$
So,
$ mv = (q)(B)(r) = (2 \times 1.6 \times 10^{-19}) \times 0.25 \times \bigl(0.83 \times 10^{-2}\bigr) \,.
$
Step 5: Calculate the de Broglie Wavelength
Planck’s constant $h = 6.6 \times 10^{-34}\,\text{J}\cdot\text{s}$. Therefore,
$ \lambda = \frac{6.6 \times 10^{-34}}{(0.83 \times 10^{-2}) \times 0.25 \times 2 \times 1.6 \times 10^{-19}} \,.
$
Carrying out this calculation gives:
$ \lambda = 0.01\,\mathring{A} \,.
$
Step 6: State the Final Answer
The de Broglie wavelength associated with the α-particle is:
0.01 Å
