Question

If the de Broglie wavelengths of deuteron, positron and electron are in the ratio $1:2:3$, then the ratio of their respective momenta is

• A. $1:2:3$
• B. $\sqrt{3}:2\sqrt{2}:1$
• C. $6:3:2$
• D. $3:2:1$
• E. $1:4:9$

Hint:

Physics Tip: Larger wavelength means smaller momentum. Just invert the wavelength ratio.

Answer 1

The Correct Option is C

Concept:
According to de Broglie relation: $$\lambda=\frac{h}{p}$$ Thus momentum is inversely proportional to wavelength: $$p\propto \frac{1}{\lambda}$$
Step 1: Given wavelength ratio.
For deuteron : positron : electron $$\lambda_d:\lambda_p:\lambda_e=1:2:3$$
Step 2: Take reciprocal ratio for momentum.
$$p_d:p_p:p_e=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}$$ Multiply by LCM $6$: $$p_d:p_p:p_e=6:3:2$$
Step 3: Final answer.
Hence correct option is (C). :contentReference[oaicite:0]{index=0}