The ratio between the de Broglie wavelength associated with protons, accelerated through a potential of 512 V and that of alpha particles accelerated through a potential of X volts is found to be one. Find the value of X.

Explanation

The accelerating potential ov V `= 512 V`
The accelerating of `alpha-`particle = X volt
Mass of the proton `= m _(p)`
Mass of the `alpha -` particle `= 4 m_(p)`
Charge of the proton `= e `
Charge of the `alpha `particle `= 2e `
de Broglie wavelength of proton `= lambda_(p)`
de Broglie wavelength of paritcle `= lambda_(a)`
`lambda - ( h )/( sqrt( 2meV))`
Ratio of Broglie wavelength associated with protons and `alpha -` particles `( lambda_(p))/( lambda_(a)) = 1 `
Formula `:`
`lambda_(p) = ( h )/( sqrt( 2m_(p) e xx 512 ))`
`lambda= ( h )/( sqrt( 2 xx 4 m_(p) xx 2eX))`
`( lambda_(p))/( lambda_(a)) =( sqrt(16 m_(p) eX))/( sqrt( 2m_(P) e 512)) = sqrt((8X)/(512))`
`:. l = sqrt(( 8X)/( 512))` ( or )` sqrt( 512 ) = sqrt( 8X) rArr 512 = 8X`
`:. X = ( 512)/( 8) =64 V`.