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प्रश्न
A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of their accelerating potentials
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उत्तर
De-broglie wavelength of the particle is given by,
`lambda = "h"/"p" = "h"/"mv" = "h"/sqrt(2"mqV")` ;
where, V= Accelerating potential and v is the speed of the particle.
Given that, the de-broglie wavelength is same for both proton and a-particle.
Charge on α particle = 2c
Mass of α -particle = 4mp
Charge on proton = qp ;
Mass of α -particle = mP
`lambda_a = lambda_P`
`=> "h"/sqrt(2"m"_alpha"q"_alpha"V"_alpha) = "h"/sqrt(2 "m"_"P" "q"_"P" "V"_"P")`
`=> "m"_alpha"q"_alpha"V"_alpha = "m"_"P" "q"_"P" "V"_"P"`
`=> "V"_"P"/"V"_alpha = ("m"_alpha "q"_alpha)/("m"_"P" "q"_"P") = (4 "m"_"P")/"m"_"P" xx (2"q"_"P")/"q"_"P" = 2/1`
2 : 1 is the required ratio of the accelerating potential.
Also,
`lambda_"a" = lambda_"p"`
`=> "h"/("m"_alpha "v"_alpha) = "h"/("m"_"P" "V"_"P")`
`=> "V"_"P"/"V"_alpha = "m"_alpha/"m"_"p" = (4"m"_"p")/"m"_"p" = 4/1`
4 : 1 is the required ratio of the speed of proton to speed of alpha-particle.
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