Question

What must be the velocity of a beam of electrons if they are to display a de-Broglie wavelength of ${{100}^{\circ }}A$ ?

Answer 1

Hint: It should be known that de-Broglie wavelength is an important concept while studying quantum mechanics. The wavelength ($\lambda$) which is associated with an object in relation to its momentum and mass is referred to as de-Broglie’s wavelength.

Formula used:
We will use the following relation:-
$\lambda =\dfrac{h}{p}$
where,
h = Planck's constant = $6.63\times {{10}^{-34}}Js$
p = momentum

Complete answer:
As we know that matter has a dual nature of waves as well as particles. de-Broglie’s wave is the characteristic of a material object that varies in space or time while behaving similar to waves. It is also known as matter-waves.
-Generally the objects which we see in our day-to-day life have wavelengths which are very small and invisible, due to which we do not experience them as waves. But in the case of subatomic particles de-Broglie wavelengths are quite visible.
-Calculation of velocity of a beam of electrons:-
We will use the following relation:-
$\lambda =\dfrac{h}{p}$
where,
h = Planck's constant = $6.63\times {{10}^{-34}}Js$
p = momentum = mv (where m is the mass and v is the velocity)
Hence we get: $\lambda =\dfrac{h}{mv}$
The values are given in the question are as follows:-
$\lambda$= ${{100}^{\circ }}A=100\times {{10}^{-10}}m={{10}^{-8}}m$
h = Planck's constant = $6.63\times {{10}^{-34}}Js$
m = mass of an electron =$9.1\times {{10}^{-31}}kg$
On substituting all these values in the above formula, we get:-
$\begin{align}
 & \Rightarrow \lambda =\dfrac{h}{mv} \\
 & \text{Rearrange the formula to obtain velocity} \\
 & \Rightarrow v=\dfrac{h}{m\lambda } \\
 & \Rightarrow v=\dfrac{6.63\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-8}}} \\
 & \Rightarrow v=7.28\times {{10}^{4}}m/s \\
\end{align}$
-Hence the velocity of a beam of electrons is $=7.28\times {{10}^{4}}m/s$.

Note:
Kindly remember to accordingly convert units and use them in the formula as required to get the correct result.
-In case of electrons going in a circle around the nuclei, the de-Broglie’s waves exist as a closed loop, such a way that they can exist only as standing waves and fit evenly around the loop. Due to this requirement, the electrons circle the nucleus in particular states or configurations which are also called stationary orbits.