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Question: The de Broglie wavelength relates to applied voltage as: A. \[\lambda = \dfrac{{12.3}}{{\sqrt h }}...

The de Broglie wavelength relates to applied voltage as:
A. λ=12.3hA0\lambda = \dfrac{{12.3}}{{\sqrt h }}{A^0}
B. λ=12.3VA0\lambda = \dfrac{{12.3}}{{\sqrt V }}{A^0}
C. λ=12.3EA0\lambda = \dfrac{{12.3}}{{\sqrt E }}{A^0}
D. None of these

Explanation

Solution

An electron that is accelerated from rest by an electric potential difference of V has a de Broglie wavelength of λ. But since the kinetic energy of the electron is equal to the energy gained from accelerating through the electric potential, λ∝1V\lambda \propto \dfrac{1}{{\sqrt V }}.

Complete step-by-step answer:
According to de Broglie, every moving particle can act as either wave or particle at different times. The wave associated with moving particles is called de Broglie wave, having de Broglie wavelength. For an electron, the de Broglie wavelength equation is written as:
λ=hmv\lambda = \dfrac{h}{{mv}}
Where, λ\lambda is the wavelength of an electron, h is Planck’s constant,
m and v are mass and velocity of an electron respectively and together it is momentum.
This states that all the moving particles have wave-particle dual nature which is true for all the particles. De Broglie wavelength of an electron can be derived in following way:
Let us suppose an electron is accelerated from rest position through a voltage difference of V volts, then the gain in kinetic energy is 12mv2\dfrac{1}{2}m{v^2} and work done on the electron or potential energy can be described as eV.
So, we can say that gain in kinetic energy is equal to work done on the electron or potential energy of an electron just before colliding with the target atom.
12mv2=eV or  V=2eVm\dfrac{1}{2}m{v^2} = eV\,or\,\,V = \sqrt {\dfrac{{2eV}}{m}}
Since, λ=hmv=h2meV\lambda = \dfrac{h}{{mv}} = \dfrac{h}{{\sqrt {2meV} }}
We can substitute the values in the above formula like
h = 6.626×10−34Js6.626 \times {10^{ - 34}}Js
m = 9.1×10−31kg9.1 \times {10^{ - 31}}kg
e = 1.6×10−19C1.6 \times {10^{ - 19}}C
This gives,λ=6.626×10−342×9×10−31×1.6×10−19×V\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times V} }}
On solving it completely, we get
λ=12.3VA0\lambda = \dfrac{{12.3}}{{\sqrt V }}{A^0}

Hence, the correct option is (B).

Note: For Objects with large mass, the wavelengths associated with ordinary objects are so short in length that their wave properties cannot be easily detected. On the other side, the wavelengths associated with electrons and other subatomic particles can be detected by experiments.