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Two particle 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

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a

1.0 × 10^-8m

b

1.25 × 10^-7 m

c

1.0 × 10^-7m

d

2.5 × 10^-8m

answer is D.

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Detailed Solution

$λ_{A} = 33.3 n m = 33.33 \times 10^{- 9} m P_{B} = \frac{1}{3} P_{A} λ_{A} = \frac{h}{P_{A}} - - - - - - - (i) λ_{B} = \frac{h}{P_{B}} = \frac{h}{P_{A} / 3} = \frac{3 h}{P_{A}} - - - - - - (i i) \frac{λ_{A}}{λ_{B}} = \frac{1}{3} λ_{B} = 3 \times λ_{A} = 3 \times 33.33 \times 10^{- 9} = 1.0 \times 10^{- 7} m$

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