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{aligned} \lambda_{A} &=33.33 nm =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}} \ldots(\text { (i) }\\ \lambda_{B} &=\frac{h}{\rho_{B}}=\frac{h}{p_{A} / 3}=\frac{3 h}{p_{A}} \ldots . .(\text { ii }) \end{aligned} $
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} $