Question #43dbd

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Answer 1 · 2017-04-22T16:51:19

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

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.

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

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

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

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