Mathematical and Physical Journal
for High Schools
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# Problem P. 4346. (April 2011)

P. 4346. A hundred years ago in 1911 Rutherford wrote in his article, that the speed of an  particle, which is moving towards the centre of a gold atom, decreases to zero at that distance, measured from the centre of the gold atom, where its electric potential energy is equal to its initial kinetic energy. The electric potential energy is partly due to the positive charge at the centre of the atom, and partly due to the negative charge which is uniformly distributed in the atom. The radius of the gold nucleus is 10-10 m.

What is this particular distance, if the speed of the particle is 2.107 m/s?

(5 pont)

Deadline expired on May 10, 2011.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A kérdéses távolság

$\displaystyle d\approx 2,7\cdot10^{-14}~\rm m.$

### Statistics:

 26 students sent a solution. 5 points: Béres Bertold, Jenei Márk, Koncz Gabriella, Kovács 444 Áron, Laczkó Zoltán Balázs, Szabó 928 Attila, Szélig Áron, Szigeti Bertalan György. 4 points: Bolgár Dániel, Filep Gábor, Pázmán Koppány. 3 points: 12 students. 2 points: 1 student. 1 point: 2 students.

Problems in Physics of KöMaL, April 2011