Dear Student,Debroglie wavelength:λ=h2mqVHere,m is mass of the particle,q is charge of the particle,V is potential differenceFor a given potential difference,λ α1mqWe have mass of alpha particle as,malpha=4×mprotonAnd charge as,qalpha=2×qprotonThe ratio of wavelength is,λprotonλalpha=malphaqalphamprotonqproton =4mproton2qprotonmprotonqproton =221Ans:Option 4)22 : 1Regards

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D e a r S t u d e n t, Debroglie wavelength : λ = \frac{h}{\sqrt{2 mqV}} Here, m is mass of the particle, q is charge of the particle, V is potential difference For a given potential difference, λ α \frac{1}{\sqrt{mq}} We have mass of alpha particle as, m_{alpha} = 4 \times m_{proton} And charge as, q_{alpha} = 2 \times q_{proton} The ratio of wavelength is, \frac{λ_{p r o t o n}}{λ_{a l p h a}} = \sqrt{\frac{m_{a l p h a} q_{a l p h a}}{m_{p r o t o n} q_{p r o t o n}}} = \sqrt{\frac{(4 m_{p r o t o n}) (2 q_{p r o t o n})}{(m_{p r o t o n}) (q_{p r o t o n})}} = \frac{2 \sqrt{2}}{1} A n s : O p t i o n 4) 2 \sqrt{2} : 1 R e g a r d s

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