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# The de-Broglie wavelength of a proton (mass = $1.6 \times {10^{ - 27}}kg$ ) accelerated through a potential difference of $1kV$ is: (A) $600{A^\circ }$ (B) $0.9 \times {10^{ - 12}}m$ (C) $7{A^\circ }$ (D) $0.9 \times {10^{ - 19}}nm$

Last updated date: 23rd Jun 2024
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Hint: We will use the de-Broglie wave equation. Firstly, we will calculate the energy of the particle. Then by substituting the values of Planck’s constant, the mass of the proton, and energy in the de-Broglie wavelength formula we will get the required answer.

Formula Used:
$\Rightarrow \lambda = \dfrac{h}{{\sqrt {2mE} }}$
$\Rightarrow E = q\Delta V$

Complete step by step solution
We know the equation derived by the scientist Louis de-Broglie which describes the wave nature of any particle. Therefore, the wavelength of any moving body/object is derived by a formula: $\lambda = \dfrac{h}{{\sqrt {2mE} }}$ ............. $\left( 1 \right)$
where $\lambda$ denotes the wavelength of a moving body, $h$ is Planck’s constant, $m$ is the mass of the particle in kg, and $E$ is the energy of the particle.
In the question we have given the potential difference as $1000V$ , that is $\Delta V = 1000V$
Now, the Energy of the particle is known by:
$\Rightarrow E = q\Delta V$ ............. $\left( 2 \right)$
We know that charge on one proton is equal to $1.6 \times {10^{ - 19}}C$ , so substituting this in the equation $\left( 2 \right)$
We get,
$\Rightarrow E = 1.6 \times {10^{ - 19}} \times {10^3}$
$\Rightarrow E = 1.6 \times {10^{ - 16}}J$ ........... $\left( 3 \right)$
And momentum, $p = mv = \sqrt {2mE}$ ............ $\left( 4 \right)$
We know that mass of a proton is $m = 1.6 \times {10^{ - 27}}kg$ ,
$\Rightarrow E = 1.6 \times {10^{ - 16}}J$ and
$\Rightarrow h = 6.62 \times {10^{ - 34}}Joule - \sec$
Substituting the above values in the equation $\left( 1 \right)$ we get,
$\Rightarrow \lambda = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 1.6 \times {{10}^{ - 27}} \times 1.6 \times {{10}^{ - 16}}} }}$
$\Rightarrow \lambda = 0.9 \times {10^{ - 12}}m$
So, the de-Broglie wavelength of a proton when accelerated through a potential difference of $1kV$ will be $0.9 \times {10^{ - 12}}m$ .
Hence, the correct answer is option (B) $0.9 \times {10^{ - 12}}m$ .