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

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Last updated date: 25th Jul 2024
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Answer
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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.