Question

The de Broglie wavelength of an electron accelerated by an electric field of V volts is given by:
A. \[\lambda {\text{ = }}\dfrac{{1.23}}{{\sqrt {\text{m}} }}\]
B. \[\lambda {\text{ = }}\dfrac{{1.23}}{{\sqrt {\text{h}} }}{\text{ m}}\]
C. \[\lambda {\text{ = }}\dfrac{{1.23}}{{\sqrt {\text{V}} }}{\text{ nm}}\]
D. \[\lambda {\text{ = }}\dfrac{{1.23}}{{\text{V}}}{\text{ }}\]

Answer 1

Hint: We know that an electron accelerating in an electric field of V volts has Kinetic energy equal to half of the products of its mass and acceleration and it is also given as the product of its charge and V. On equating these two we can get the value of wavelength.

Complete step by step answer:
For an electron accelerated by an electric field V is eV. But we know its kinetic energy is also given by –
\[{\text{K}}{\text{.E = }}\dfrac{1}{2}{\text{ m}}{{\text{u}}^2}\]
\[\dfrac{1}{2}{\text{ m}}{{\text{u}}^2}{\text{ = eV}}\]

We can then rearrange as - \[{{\text{u}}^{\text{2}}}{\text{ = }}\dfrac{{{\text{2eV}}}}{{\text{m}}}\]
Therefore, \[{\text{u}}{\text{ = }}\sqrt {\dfrac{{{\text{2eV}}}}{{\text{m}}}} \] -----(1)

We know that the De Broglie wavelength is given by - \[\text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\text{mu}}\] ---(2)
From (1) and (2) we can see – \[\text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\text{m }\!\!\times\!\!\text{ }\sqrt{\dfrac{\text{2eV}}{\text{m}}}}\]
                                                              \[\text{ }\!\!\lambda\!\!\text{ = }\dfrac{\text{h}}{\sqrt{\text{2eVm}}}\] -------(3)

We know that the value of Planck's constant ‘h’ is given by \[6.626{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 34}}{\text{ Js}}\].
Charge on an electron (e) is given by \[\text{1}\text{.6 }\times \text{ 1}{{\text{0}}^{-19}}\text{ C}\] and its mass (m) is given as \[\text{9}\text{.1 }\times \text{ 1}{{\text{0}}^{-31}}\text{ Kg}\].

Substituting all these values in equation number (3) we get,
\[\lambda \,=\text{ }\dfrac{6.626\text{ }\times \text{ 1}{{\text{0}}^{-34}}\text{ Js}}{\sqrt{2\ \times \text{ 1}\text{.6 }\times \text{ 1}{{\text{0}}^{-19}}\ \text{C }\times \text{ V }\times \text{ 9}\text{.1 }\times \text{ 1}{{\text{0}}^{-31}}\text{ Kg}}}\]

On solving we get, \[\lambda \text{ = }\dfrac{\text{1}\text{.23}\ \text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}}{\sqrt{\text{V}}}\text{ m}\]
But we know that, \[\text{1nm = 1}{{\text{0}}^{-9}}\text{m}\]. So, \[\lambda \text{ = }\dfrac{\text{1}\text{.23}\ }{\sqrt{\text{V}}}\text{ nm}\]

Hence, option C is correct.

Additional information: The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. De Broglie wavelength is usually inversely proportional to its force. We know that electrons revolve in circles around the nucleus and hence their De Broglie wavelength exists in the form of a closed loop.

Note: The wave properties of matter are only observable for very small i.e. microscopic objects and not macroscopic systems.