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Question

Question: If the kinetic energy of the particle is increased to \[16\] times its previous value, the percentag...

If the kinetic energy of the particle is increased to 1616 times its previous value, the percentage change in the de Broglie wavelength of the particle is
A). 2525
B). 7575
C). 6060
D). 5050

Explanation

Solution

From the de-Broglie wavelength formula we can see that wavelength is inversely related to its momentum. We have the equation which relates momentum and the kinetic energy of a particle. Here it is asked to find out the percentage change in wavelength if we increase the kinetic energy. By substituting the equation for momentum in the de Broglie equation we can determine the percentage change in wavelength.
Formula used:
λ=hp\lambda =\dfrac{h}{p}
Kinetic energy = p22m\text{Kinetic energy = }\dfrac{{{p}^{\text{2}}}}{\text{2m}}

Complete step-by-step solution:
De Broglie wavelength is given by,
λ=hp\lambda =\dfrac{h}{p}---------(1)
Where,
hh is the Planck’s constant
pp is the momentum of the particle
We know that,
p=mvp=mv
Where,
mm is the mass of the particle
vv is the velocity of the particle
Also we have,
Kinetic energy = p22m\text{Kinetic energy = }\dfrac{{{p}^{\text{2}}}}{\text{2m}}
Then,
p=2mKEp=\sqrt{2mKE} --------(2)
Substituting 2 in equation 1, we get,
λ=h2mKE\lambda =\dfrac{h}{\sqrt{2mKE}}
From the above equation we can see that de Broglie wavelength is inversely proportional to the kinetic energy.
If we increase kinetic energy, the new de Broglie wavelength can be given as,
λ1=h2mK.E1{{\lambda }_{1}}=\dfrac{h}{\sqrt{2mK.{{E}_{1}}}} ----------(3)
Where,
K.E1K.{{E}_{1}} is the change in kinetic energy.
The new de Broglie wavelength when kinetic energy is increased 16 times,
λ1=h2m×16K.E=h42mKE=λ4{{\lambda}_{1}}=\dfrac{h}{\sqrt{2m\times16K.E}}=\dfrac{h}{4\sqrt{2mKE}}=\dfrac{\lambda }{4}
λ1=λ4\Rightarrow {{\lambda }_{1}}=\dfrac{\lambda }{4}
Then,
Percentage change in de Broglie wavelength of the particle =λλ4λ×100=34×100=75=\dfrac{\lambda -\dfrac{\lambda }{4}}{\lambda }\times 100=\dfrac{3}{4}\times 100=75%
The answer is option A.

Note: In gases, liquids, and solids increase in kinetic energy has different effects. In gases, if the kinetic energy is increased, particles move faster in all directions and collide with each other. In the case of liquids, increased kinetic energy results in an increased collision rate. Hence the diffusion rate also increases. In solids, particles are packed tightly as possible. When kinetic energy is increased, the particles start vibrating.