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 $d$ between the atoms of the array is $2 \,\mathring{A}$. A similar standing wave is again formed, if $d$ is increased to $2.5 \,\mathring{A}$ but not for any intermediate value of $d$.
The energy of the electron is

• A. $160 \,eV$
• B. $150.8 \,eV$
• C. $145 \,eV$
• D. $100 \, eV$

Answer 1

Answer: (B)

de-Broglie wavelength is given by
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 K m}}$
Here, $K=$ Kinetic energy of electron
$\therefore \quad K=\frac{h^{2}}{2 m \lambda^{2}}=\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2\left(9.1 \times 10^{-31}\right)\left(10^{-10}\right)^{2}}$
$=2.415 \times 10^{-17}\, J$
$K=\left(\frac{2.415 \times 10^{-17}}{1.6 \times 10^{-19}}\right)=150.8\, eV$