Question

Two particles A and B are in motion and the wavelength of A is 5 × 10-8 m. The wavelength of B, so that its momentum is double that of A, will be:

• A. (A) 3 × 10-4 m
• B. (B) 2.5 × 10-8 m
• C. (C) 2.5 × 10-6 m
• D. (D) 1.6 × 10-8 m

Answer 1

Answer: (B)

(B) 2.5 × 10-8 m

Answer 2

Explanation:
Given:Wavelength of A, λA = 5 × 10-8 mMomentum of B = 2 × Momentum of AAccording to de-Broglie's equation : λ = hp ..........(i)where, λ = Wavelengthh = Planck's constantp = MomentumThus, for particle A:λA=hPA .........(ii)For particle B:λB=hpB .........(iii)As, PB=2×pA, we we have from equationλB=h2pA ........(iv)Dividing equation (ii) by equation (iv), we getλAλB=hPAh2PAλAλB=2λB=λA2=5×10−82=2.5×10−8Hence, the correct option is (B).