Advertisement Remove all ads

A Proton and an α-particle Have the Same De-broglie Wavelength. Determine the Ratio of Their Accelerating Potentials - Physics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

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

Advertisement Remove all ads

Solution

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.  

Concept: de-Broglie Relation
  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×