Question

The de-Broglie wavelength associated with the electron in the n=4 level is:
A. half of the de-Broglie wavelength of the electron in the ground state
B. Four times the de-Broglie wavelength of the electron in the ground state
C. $\dfrac{1}{4}th$ of the de-Broglie wavelength of the electron in the ground state
D. Two times the de-Broglie wavelength of the electron in the ground state

Answer 1

Firstly recall the expression for de Broglie wavelength or you could derive it using Einstein’s mass-energy relation and Planck’s equation. Rearranging the expression of angular momentum you could find an expression for linear momentum and then you could substitute this in the expression for de Broglie wavelength. Also substitute for the radius of nth in the expression and then find the de Broglie wavelength for n=1 (ground state) and n=4 and find their ratio to get the answer.

Formula used:
Expression for de Broglie wavelength,
$\lambda =\dfrac{h}{mv}$
Expression for angular momentum,
$mvr=\dfrac{nh}{2\pi }$
Expression for radius of nth orbit of an atom,
${{r}_{n}}={{r}_{0}}\dfrac{{{n}^{2}}}{Z}$

Complete step-by-step answer:
We know that the de-Broglie wavelength 𝛌 associated with an object is directly related to its momentum and mass. Using some of the well established theories the French scientist Louis de Broglie derived the expression for de Broglie wavelength.
From Einstein’s equation relating matter and energy we have,
$E=m{{c}^{2}}$ …………………… (1)
Where, E= energy, m=mass, c=speed of light
Planck’s theory states that every quantum of a wave consists of a discrete amount of energy which can be given by the Planck’s equation as,
$E=h\nu$ …………………….. (2)
Where,
E = energy
h = Planck’s constant
$\nu$ = frequency
As per de Broglie’s belief particles and waves show the same characteristics, and hence he hypothesized that the above two energies would be equal. Therefore,
$\Rightarrow m{{c}^{2}}=h\nu$
Since, the real particles do not travel at the speed of light c, c could be replaced by velocity v.
$\Rightarrow m{{v}^{2}}=h\nu$ ………………… (3)
But we know a relation that relates frequency, velocity and wavelength as,
$\nu =\dfrac{c}{\lambda }$
Substituting this in (3) we get,
$\Rightarrow m{{v}^{2}}=\dfrac{hv}{\lambda }$
$\Rightarrow \lambda =\dfrac{h}{mv}$ ……………………… (4)
Equation (4) is the expression for de Broglie wavelength.
We also have the expression for angular momentum as,
$mvr=\dfrac{nh}{2\pi }$
$\Rightarrow mv=\dfrac{nh}{2\pi r}$
Substituting in equation (4) we get,
$\lambda =\dfrac{h}{\dfrac{nh}{2\pi r}}=\dfrac{2\pi r}{n}$
$\Rightarrow n\lambda =2\pi r$ ………………………. (5)
We also have the expression for nth orbit of an atom with atomic number Z as,
${{r}_{n}}={{r}_{0}}\dfrac{{{n}^{2}}}{Z}$
Substituting this in (5) we get,
$\Rightarrow n\lambda =2\pi {{r}_{0}}\dfrac{{{n}^{2}}}{Z}$
$\Rightarrow \lambda =\dfrac{2\pi {{r}_{0}}n}{Z}$ ……………………….. (6)
Substituting n=1 in (6) we get de Broglie wavelength associated with the ground state as,
${{\lambda }_{n=1}}=\dfrac{2\pi {{r}_{0}}}{Z}$ ………………………. (7)
For the 4th level (n=4),
${{\lambda }_{n=4}}=\dfrac{2\pi {{r}_{0}}\left( 4 \right)}{Z}$ …………………….. (8)
Dividing (8) by (7) we get,
$\dfrac{{{\lambda }_{n=4}}}{{{\lambda }_{n=1}}}=\dfrac{2\pi {{r}_{0}}\left( 4 \right)}{Z}\times \dfrac{Z}{2\pi {{r}_{0}}}$
$\Rightarrow {{\lambda }_{n=4}}=4{{\lambda }_{n=1}}$
Therefore, the de-Broglie wavelength associated with the electron in the n=4 level is four times the de-Broglie wavelength of the electron in the ground state.

So, the correct answer is “Option B”.

Note: At times it is better to memorize the standard results for certain quantities. Otherwise, you will have to derive them which will surely consume time. Even if we were to derive the de Broglie wavelength, you had to know Einstein’s mass-energy relation and Planck’s equation. We have also used certain other standard expressions like that for the radius of nth orbit, expression for angular momentum, etc.