Question

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

Answer 1

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
$$\lambda ={\displaystyle \frac{h}{mv}}$$
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
$$\lambda ={\displaystyle \frac{h}{mv}}$$
Where, $$m$$is the mass of that particle, $$v$$is the velocity of that particle.
Now, this can also be written as
$$\lambda ={\displaystyle \frac{h}{p}}$$
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
$$\lambda ={\displaystyle \frac{h}{\sqrt{4meV}}}$$
Now,
$$\begin{gathered} h=6.626\times {10}^{-34}Js \\ m=4\times 1.6\times {10}^{-27}Kg \\ e=1.6\times {10}^{-19}C \end{gathered}$$

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

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.