Question

Two particles A and B are in motion. If the wavelength associated with ‘A’ is 33.33 nm, the wavelength associated with ‘B’ whose momentum is 1/3 rd of ‘A ’ is

• A. 1.25 x 10⁻⁷ m
• B. 1.0 x 10⁻⁷ m
• C. 1.0 x 10⁻⁸ m
• D. 2.5 x 10⁻⁸ m

Answer 1

Answer: (B)

$\begin{array}{rl}{\lambda }_A& =33.33nm=33.33\times 10^{-9}m\\ p_B& =\frac{1}{3}{p}_A\\ {\lambda }_A& =\frac{h}{p_A}\\ {\lambda }_A& =\frac{h}{p_A}\dots (\text{ (i) }\\ {\lambda }_B& =\frac{h}{{\rho }_B}=\frac{h}{p_A/3}=\frac{3h}{p_A}\dots ..(\text{ ii })\end{array}$
From eq. (i) and (ii)
$\begin{array}{l}\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{3}\\ \therefore {\lambda }_B=3\times {\lambda }_A=3\times 33.33\times 10^{-9}m\\ =1.0\times 10^{-7}m\end{array}$