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If the kinetic energy of the particle is increased to $16$ times its previous value, the percentage variation in the de-Broglie wavelength of the particle will be,
  & A.25 \\
 & B.75 \\
 & C.60 \\
 & D.50 \\

Last updated date: 13th Jun 2024
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Hint: The wavelength of the particle will be the ratio of the Planck’s constant to the square root of the product of twice the mass and the kinetic energy. Using this relationship, find the wavelength in the first case and then find the wavelength if the energy increases. Find out the percentage variation using this. These details will help you in solving this question.

Complete step by step solution:
The de Broglie wavelength is given by the equation,
$\lambda =\dfrac{h}{p}$
Where $h$ be the Planck’s constant and $p$ be the momentum of the particle.
The momentum can be written as,$p=\sqrt{2mE}$
Substituting this in the equation will give,
$\lambda =\dfrac{h}{\sqrt{2mE}}$
Where $m$ be the mass of the particle, $E$ be the kinetic energy of the particle.
The kinetic energy given in the first case is $E$ itself. Therefore the equation will be,
$\lambda =\dfrac{h}{\sqrt{2mE}}$
Squaring this will give,
${{\lambda }^{2}}=\dfrac{{{h}^{2}}}{2mE}$
Now let us look at the second case in the question. In this case the kinetic energy has been increased by $16$ times its previous value. That is,
Therefore, the wavelength of the particle can be written by the equation,
${{{\lambda }'}^{2}}=\dfrac{{{h}^{2}}}{2m\left( 16E \right)}$
From this we can see that, the square of the wavelength has become ${{\dfrac{1}{16}}^{th}}$ of the square of the wavelength of the previous case. That is,
${{{\lambda }'}^{2}}=\dfrac{{{\lambda }^{2}}}{16}$
Taking the square root will give,
${\lambda }'=\dfrac{\lambda }{4}$
The change in the wavelength can be written as,
  & \Delta \lambda =\lambda -{\lambda }' \\
 & \Rightarrow \Delta \lambda =\lambda -\dfrac{\lambda }{4} \\
 & \Rightarrow \Delta \lambda =\dfrac{3\lambda }{4} \\
The percentage change in the wavelength can be written as,
$\text{percentage change=}\dfrac{\text{change in }\lambda }{\text{original }\lambda }$
That is,
$\text{percentage change}=\dfrac{\Delta \lambda }{\lambda }\times 100$
Substituting the values will give,
$\text{percentage change}=\dfrac{3}{4}\times 100=75%$

The correct answer is given as option B.

Kinetic energy is the energy acquired by the body because of its motion. This is a scalar quantity also. De Broglie wavelength is a wavelength which determines the probability density of calculating the position of the object at a specific point of the configuration space.