Question #43dbd

1 Answer
Apr 22, 2017

Answer:

#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)))#