a) When a proton-beam approaches the nucleus of a gold-atom, it is scattered at

Question

a) When a proton-beam approaches the nucleus of a gold-atom, it is scattered at large angles. The kinetic energy of the proton beam is 7.7 MeV. If its closest approach is Rp; estimate the size of the nucleus. The atomic number of Au nucleus Z=79 and remember that the proton has a charge q=+e. Besides the calculations, you may briefly discuss your logical approach.

b) The transition from first excited state to the ground state of K (potassium) results in emission of a photon =770 nm if the K vapors are used in Franck-Hertz experiment, at what accelerating voltage you would find the first dip (decrease) in current?

Q.3: (a) When a proton-beam approaches the nucleus of a gold-atom, it is scattered at large angles. The kinetic energy of the proton beam is 7.7 Mev. If its closest approach is Rp: Estimate the size of the nucleus. The atomic number of Au nucleus Z-79 and remember that the proton has a charge q-+e. Besides the calculations, you may briefly discuss your logical approach

Explanation / Answer

Solution:-

a) Here we have to equate the kinetic energy of @ with the colulomb PE (potential energy) of @+nucleus (here we have to assume non relativistic kinematics).

So here at distance f, the d0 which is thw closest approach, KE (kinetic energy) of incident @ completely converted to PE (potential energy).

Au: T= q1q2/4o

Therefore do =(q1q2/40T)

= ( (2e)(79e) / (4(8.85419x10-12)(7.7x106)e) )

= 29.5 fm

Hence the size of nucleus in terms of do is 29.5 fm.

b) Delta E = hc/ = 1240/770 = 1.610 eV.

The first decrease in current shall occur when the voltage reaches 1.610 eV.

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