00:01
In this problem on the topic of elementary particle physics, we are told that the debrugly wavelength lambda of an alpha particle with energy 5 mega electron volts is 6 .4 femtometers.
00:11
The closest distance armine is the distance to the gold nucleus that the alpha particle can get, and this is 45 .5 femtometers.
00:21
We want to know how the ratio armin over lambda will vary with the kinetic energy of the alpha particle.
00:27
Now the potential energy at the closest approach, u at r min, is equal to the kinetic energy of the alpha particle initially, which is the electric constant k times the proton number for the alpha particle times the charge number for the gold nucleus times e squared over the distance of closest approach r min.
00:59
Rearranging, we get armin to be little k times x alpha times z -a -u times e -squared divided by k, which means that armin over lambda is equal to k -z -alpha, z -a -u, e -squared p divided by k -h, where lambda is now the dibrogly wavelength.
01:40
And this is equal to k -z -alpha, z -a -u, times e -squared, times the mass of the alpha particle times its speed v, divided by k -h...