Question

Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance ${}^{\prime }{d}^{\prime }$ between the atoms of the array is $2\mathring{A}$. A similar standing wave is again formed if ${}^{\prime }{d}^{\prime }$ is increased to $2.5\mathring{A}$ but not for any intermediate value of $d$.
Find the energy of the electrons in $eV$:

• A. $302eV$
• B. $151eV$
• C. $75.5eV$
• D. $75.5\times {10}^{6}eV$

Answer 1

The correct option is B $151eV$

We know that,

The interatomic spacing depends upon the path difference.

Given that,

$d_1=2\stackrel{\circ }{A}$

$d_2=2.5\stackrel{\circ }{A}$

We know, $d=\frac{n\lambda }{2}$

So,

$d_1=\frac{n\lambda }{2}$

$d_2=\frac{(n+1)\lambda }{2}$

Now, the interatomic spacing is

$d_2-d_1=\frac{\lambda }{2}$

$\lambda =2(d_2-d_1)$

Now, put the value of d₁ and d₂

Then,

$\lambda =2(2.5-2)$

$\lambda =1\stackrel{\circ }{A}$

We know that,

$\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

Now, energy of the electron

$E=\frac{1}{2}m{v}^2$

$E=\frac{1}{2m}\times m^2{v}^2$

We know that,

$p=mv$

So,

$E=\frac{p^2}{2m}$

$E=\frac{h^2}{2m{\lambda }^2}$

$E=\frac{{(6.6\times {10}^{-34})}^2}{2\times 9.1\times {10}^{-31}\times {\left({10}^{-10}\right)}^2}$

$E=150.95eV$

$E=151eV$

Hence, the energy of the electron is $151$eV.