Question

If the kinetic energy of a particle is doubled, de Broglie wavelength becomes:
A. 2 times
B. 4 times
C. $\sqrt{2}$times
D. ${\displaystyle \frac{1}{\sqrt{2}}}$times

Answer 1

Hint:Matter waves, which are an example of wave–particle duality, are an important element of quantum mechanics theory. All stuff behaves in a wavelike manner. A beam of electrons, for example, can be diffracted in the same way as a beam of light or a water wave can. However, in most situations, the wavelength is too short to have a practical effect on daily tasks.

Complete step by step solution:
The De Broglie wavelength is a wavelength exhibited in all objects in quantum mechanics that defines the probability density of locating the item at a particular position in the configuration space, according to wave-particle duality. The momentum of a particle is inversely related to its de Broglie wavelength.
The kinetic energy of an item is the energy it has owing to its motion in physics. It is the amount of effort required to propel a body of a given mass from rest to a certain velocity. The body retains its kinetic energy after gaining it during acceleration unless its speed changes. When the body decelerates from its current speed to a condition of rest, it does the same amount of effort. The kinetic energy ${\displaystyle \frac{1}{2}}m{v}^2$ of a non-rotating object of mass m moving at a speed v is in classical mechanics. Only when v is substantially less than the speed of light is this a fair approximation in relativistic mechanics.
$\text{KE}={\displaystyle \frac{1}{2}}\text{m}{\text{v}}^2$
Hence $\text{v = }\sqrt{\text{2mKE}}$
We know that Linear momentum is defined as the product of an object's mass and velocity in Newtonian physics. It's a two-dimensional vector quantity with a magnitude and a direction. When m is the mass of an item and v is its velocity (also a vector quantity), the object's momentum is
$\text{p}=\text{mv}$
$\Rightarrow \text{p}=m\sqrt{2\text{mKE}}$
$\Rightarrow p=m\sqrt{2KE}$
We know that
$\Rightarrow \lambda ={\displaystyle \frac{\text{h}}{\text{p}}}$
If KE is doubled, p will become$\sqrt{2}$times.
$\lambda$will become${\displaystyle \frac{1}{\sqrt{2}}}$ times.
Hence option D Is correct.

Note:
The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion. Momentum varies depending on the frame of reference, but it is a conserved quantity in any inertial frame, meaning that if a closed system is not influenced by external forces, its total linear momentum remains constant. Momentum is maintained in special relativity (with a modified formula) as well as electrodynamics, quantum mechanics, quantum field theory, and general relativity (in a modified version). It is a manifestation of translational symmetry, which is one of the fundamental symmetries of space and time.