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Question: Relativistic corrections become necessary when the expression for the kinetic energy \(\dfrac{1}{2}m...

Relativistic corrections become necessary when the expression for the kinetic energy 12mv2\dfrac{1}{2}m{v^2}, becomes comparable with mc2m{c^2}, where mm is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?
(A) λ=10nm\lambda = 10nm
(B) λ=10−2nm\lambda = {10^{ - 2}}nm
(C) λ=10−4nm\lambda = {10^{ - 4}}nm
(D) λ=10−6nm\lambda = {10^{ - 6}}nm

Explanation

Solution

To solve this question, we need to use the de Broglie relation between the momentum and the wavelength of the electron. For each value of the wavelength given in the options, we have to find out the value of the speed of the electron. Then we have to compare these values of speeds with the speed of light in vacuum.
Formula used: The formula used to solve this question is given by
p=hλp = \dfrac{h}{\lambda }, here pp is the momentum of a particle through which matter waves of wavelength λ\lambda are emitted, and hh is the Planck’s constant.
p=mvp = mv, here pp it the momentum of a particle of mass mm moving with a velocity of vv.

Complete step-by-step solution:
The expression of the kinetic energy of a particle, which is given by 12mv2\dfrac{1}{2}m{v^2}, will become comparable with mc2m{c^2}, when the speed vv of the particle approaches to the speed of light in vacuum, cc.
So we have to find out the speed of the electron for each value of the de Broglie wavelength of the electron given in the options. And then compare them with the speed of light in vacuum, which is approximately equal to 3×108m/s3 \times {10^8}m/s.
We know that the de Broglie relation is given by the expression
p=hλp = \dfrac{h}{\lambda }................................(1)
Now, we know that the momentum of a particle is given by the relation
p=mvp = mv
Substituting this in (1) we have
mv=hλmv = \dfrac{h}{\lambda }
Dividing by mm we get
v=hmλv = \dfrac{h}{{m\lambda }}......................(2)
Now, we know that the value of the Planck’s constant is 6.6×10−34Js6.6 \times {10^{ - 34}}Js. Also, the mass of an electron is approximately equal to 9.1×10−31kg9.1 \times {10^{ - 31}}kg. Substituting these in (2) we get
v=6.6×10−349.1×10−31×λv = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times \lambda }}...........(3)
Now, from the option A, we have λ=10nm=10×10−9m\lambda = 10nm = 10 \times {10^{ - 9}}m. Substituting this in (3), we get
v=6.6×10−349.1×10−31×10×10−9v = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times 10 \times {{10}^{ - 9}}}}
On solving we get
v=7.25×104m/sv = 7.25 \times {10^4}m/s
This value of speed is clearly much less than 3×108m/s3 \times {10^8}m/s. So no relativistic correction is needed.
Therefore, option A is incorrect.
From option B, we have λ=10−2nm=10−2×10−9m\lambda = {10^{ - 2}}nm = {10^{ - 2}} \times {10^{ - 9}}m. Substituting this above, we get
v=6.6×10−349.1×10−31×10−2×10−9v = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times {{10}^{ - 2}} \times {{10}^{ - 9}}}}
On solving we get
⇒v=7.25×107m/s\Rightarrow v = 7.25 \times {10^7}m/s
This value of speed is not comparable to 3×108m/s3 \times {10^8}m/s. So no relativistic correction is needed.
Therefore, option B is also incorrect.
From the option C, we have λ=10−4nm=10−4×10−9m\lambda = {10^{ - 4}}nm = {10^{ - 4}} \times {10^{ - 9}}m. Substituting this above, we get
v=6.6×10−349.1×10−31×10−4×10−9v = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times {{10}^{ - 4}} \times {{10}^{ - 9}}}}
On solving we get
⇒v=7.25×109m/s\Rightarrow v = 7.25 \times {10^9}m/s
This value of speed is clearly greater than 3×108m/s3 \times {10^8}m/s. We know that the speed of any particle in the universe cannot exceed the value of the speed of light in vacuum .So relativistic correction is needed.
Therefore, option C is correct.
From the option D, we have λ=10−6nm=10−6×10−9m\lambda = {10^{ - 6}}nm = {10^{ - 6}} \times {10^{ - 9}}m. Substituting this above, we get
v=6.6×10−349.1×10−31×10−6×10−9v = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times {{10}^{ - 6}} \times {{10}^{ - 9}}}}
On solving we get
⇒v=7.25×1011m/s\Rightarrow v = 7.25 \times {10^{11}}m/s
This value of speed is clearly greater than 3×108m/s3 \times {10^8}m/s. Similar to the above case, we know the speed of any particle in the universe cannot exceed the value of the speed of light in vacuum .So relativistic correction is needed.
Therefore, option D is also correct.

Hence the correct options are C and D.

Note: We should not forget to convert the value of the wavelengths given in the options into the SI units. If you substitute them as it is without converting, then the values of the speeds obtained would be very less.