Question

Question #43dbd

Answer 1

`sf(lambda_(n)=(h)/(8sqrt(2meV))`

Explanation:

I will assume that the charge on the proton is +e and that the alpha particle is 4 times as massive.

The kinetic energy of the proton as it is accelerated through the potential V is given by:

`sf(eV=1/2mv^2)`

`:.` `sf(v^2=(2eV)/m)`

`sf(v=sqrt((2eV)/(m))" "color(red)((1)))`

The de Broglie wavelength of the proton is given by:

`sf(lambda_(p)=h/(mv)" "color(red)((2)))`

Substituting the expression for v from `sf(color(red)((1)))` into `sf(color(red)((2))rArr)`

`sf(lambda_p=h/(msqrt((2eV)/(m)))`

The mass of the alpha particle = 4m and the charge = 2e.

`:.` `sf(lambda_(n)=(h)/(4msqrt((4eV)/(m))`

`sf(lambda_(n)=h/(4m^(1/2).(4eV)^(1/2))`

`sf(lambda_(n)=h/(4xx2sqrt(meV)))`

`sf(lambda_(n)=h/(8sqrt(meV)))`