Consider a deuteron and α - particle in motion such that respectively de Broglie wavelength are λ1 and λ2 Column I gives physical quantity which is same for both the particle column II gives value of λ1λ2. Match appropriately Column I Column IIA)MomentumP)2B)speedQ)2C)Kinetic energyR)1D)Potential difference through which the particle have been accelerated S)12

λ=hpif p1=p2 then λ1λ2=1λ=hmvif υ1=υ2 then λ1λ2=m2m1=42=2 λ=h2mKEif KE1=KE2 then λ1λ2=m2m1=2 y=h2mqυif V1=V2 then λ1λ2=m2q2m1q1 λ1λ2=4×22×1=2

Consider a deuteron and α - particle in motion such that respectively de Broglie wavelength are λ1 and λ2 Column I gives physical quantity which is same for both the particle column II gives value of λ1λ2. Match appropriately Column I Column IIA)MomentumP)2B)speedQ)2C)Kinetic energyR)1D)Potential difference through which the particle have been accelerated S)12

λ=hpif p1=p2 then λ1λ2=1λ=hmvif υ1=υ2 then λ1λ2=m2m1=42=2 λ=h2mKEif KE1=KE2 then λ1λ2=m2m1=2 y=h2mqυif V1=V2 then λ1λ2=m2q2m1q1 λ1λ2=4×22×1=2

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Consider a deuteron and $α$ - particle in motion such that respectively de Broglie wavelength are $λ_{1} a n d λ_{2}$ Column I gives physical quantity which is same for both the particle column II gives value of $\frac{λ_{1}}{λ_{2}}$ . Match appropriately
Column I Column II
A) Momentum P) $\sqrt{2}$
B) speed Q) $2$
C) Kinetic energy R) 1
D) Potential difference through which the particle have been accelerated S) $\frac{1}{2}$

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A-P,B-Q,C-Q,D-S

A-Q ,B-Q,C-P,D-P

A-R,B-Q,C-P,D-Q

A-R,B-Q,C-P,D-P

answer is D.

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Detailed Solution

$λ = \frac{h}{p} i f p_{1} = p_{2} t h e n \frac{λ_{1}}{λ_{2}} = 1$

$λ = \frac{h}{m v} i f υ_{1} = υ_{2} t h e n \frac{λ_{1}}{λ_{2}} = \frac{m_{2}}{m_{1}} = \frac{4}{2} = 2 λ = \frac{h}{\sqrt{2 m K E}} i f K E_{1} = K E_{2} t h e n \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{m_{2}}{m_{1}}} = \sqrt{2}$

$y = \frac{h}{\sqrt{2 m q υ}} i f V_{1} = V_{2} t h e n \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{m_{2} q_{2}}{m_{1} q_{1}}} \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{4 \times 2}{2 \times 1}} = 2$

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