Question

A standing wave having $3$ nodes and $2$ antinodes is formed between two atoms having a distance $1.21\mathring{A}$ between them. The wavelength of the standing wave is

• A. $1.21\mathring{A}$
• B. $3.63\mathring{A}$
• C. $4.84\mathring{A}$
• D. $5.86\mathring{A}$

Answer 1

Answer: (A)

$1.21\mathring{A}$

The given situation can be shown as
Given the distance between the two atoms $=1.21\mathring{A}$
so, the wavelength of the standing wave is
$\frac{\lambda }{2}+\frac{\lambda }{2}=\lambda =1.21\mathring{A}$