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What is the de-Broglie's wavelength of the $\alpha -$particle accelerated through a potential difference$V$?

Last updated date: 25th Jul 2024
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Hint: In order to solve this question, we are going to first write the formula for the de-Broglie wavelength of any particle, then, we will relate it with the information given that is the case of an$\alpha -$particle accelerated through a potential difference$V$, then, we will put the values in the final equation.

Formula used:
The de-Broglie wavelength of any particle is given by the formula
$\lambda = \dfrac{h}{{mv}}$
Where, $m$is the mass of that particle, $v$is the velocity of that particle.

Complete step-by-step solution:
We know that the de-Broglie wavelength of any particle is given by the formula
$\lambda = \dfrac{h}{{mv}}$
Where, $m$is the mass of that particle, $v$is the velocity of that particle.
Now, this can also be written as
$\lambda = \dfrac{h}{p}$
We know that the momentum-energy relation can be expressed as
$p = \sqrt {2mE}$
Here, the$\alpha -$particle is accelerated through the potential difference of$V$volts. It is having a charge equal to $2e$
Hence, the energy becomes equal to $E = 2eV$
Thus, the momentum relation for an$\alpha -$particle becomes equal to:
$p = \sqrt {4meV}$
Thus, the de-Broglie wavelength of the$\alpha -$particle is equal to
$\lambda = \dfrac{h}{{\sqrt {4meV} }}$
Now,
$h = 6.626 \times {10^{ - 34}}Js \\ m = 4 \times 1.6 \times {10^{ - 27}}Kg \\ e = 1.6 \times {10^{ - 19}}C \\$

Putting these values in the relation, we get
$\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}Js}}{{\sqrt {4 \times 4 \times 1.6 \times {{10}^{ - 27}} \times 1.6 \times {{10}^{ - 19}}V} }}$
Solving this, we get
$\lambda = \dfrac{{1.01 \times {{10}^{ - 11}}}}{{\sqrt V }}$
This implies that the wavelength is equal to
$\lambda = \dfrac{{0.101}}{{\sqrt V }}\mathop A\limits^{\,\,\,\,\,0}$

Note: It is important to note the step where the momentum is related with the potential difference through which is applied across the $\alpha -$particle. Putting off the values keeping in mind that for the $\alpha -$particle, charge is twice that of the electron and the mass is four times that of an electron is also important.