Question

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 1

Hint: According to de- Broglie principle, matter acts as a wave same as that of light which has dual nature of both matter and particles. The de- Broglie wavelength helps to determine the idea of matter that has a wavelength. The particles can be microscopic or macroscopic.

Complete step by step solution:
If the proton has a charge ‘q’ and it is accelerated through a potential applied then the kinetic energy of the proton is given by
$E = qV$---(i)
The formula for the kinetic energy of a moving particle is given by
$E = \dfrac{1}{2}m{v^2}$---(ii)
But the momentum of a body is also the product of its mass and velocity. So it can be written as
$p = mv$
Substituting the value of momentum in equation (ii),
$E = \dfrac{{{p^2}}}{{2m}}$
$\Rightarrow p = \sqrt {2mE} $---(iii)
The formula for de- Broglie wavelength of a moving body is given by
$\Rightarrow \lambda = \dfrac{h}{p}$---(iv)
Where ‘h’ is Planck’s constant
And ‘p’ is the momentum
Substituting the value of momentum from equation (iii) to (iv),
$\Rightarrow \lambda = \dfrac{h}{{\sqrt {2mE} }}$
Substituting the value of E from equation (i),
$\Rightarrow \lambda = \dfrac{h}{{\sqrt {2meV} }}$
Substituting all the values in the above equation,
$\Rightarrow \lambda = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 1.6 \times {{10}^{ - 27}} \times 1000 \times 1.6 \times {{10}^{ - 27}}} }}$
Solving above equation, we get
$\Rightarrow \lambda = 0.9 \times {10^{ - 12}}m$

Option B is the right answer.

Note: It is important to note that the de- Broglie equation is used to define the wave properties of matter. The particle and the wave nature of the matter are complementary to each other but it is not mandatory that both of them are present at the same time. The de- Broglie equation is more useful for microscopic particles.