Question

A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of their accelerating potentials

Answer 1

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 = 4m_p

Charge on proton =  q_{p ;}

Mass of α -particle = m_P 

`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.