# 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 $

Answer

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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 $ .**

**Additional information**

The De-Broglie equation states that matter can act as a wave just like light behaves like a wave and particles. The equation also explains that a beam of electrons can be diffracted similar to a beam of light. Thus, the de-Broglie equation explains the idea of matter having a wavelength. Therefore, according to de-Broglie every moving particle whether it is microscopic or macroscopic it will have a wavelength.

**Note**

Electromagnetic potential of particles diminishes in the reverse extent of the good ways from the molecule to the perception point, the potential of solid collaboration in the gravitational examples of solid communication acts similarly. Remember the relation between potential difference and momentum.

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