Question

A beam of 7 MeV \(\alpha\) particles scatters from a thin gold foil. What is the ratio of the number of \(\alpha\) particles scattered at angles greater than 5° to the number scattered at angles greater than 25°? \frac{N_{\alpha} (\theta > 5^{\circ})}{N_{\alpha} (\theta > 25^{\circ})} = 28.4

          A beam of 7 MeV \(\alpha\) particles scatters from a thin gold foil. What is the ratio of the number of \(\alpha\) particles scattered at angles greater than 5° to the number scattered at angles greater than 25°?
\frac{N_{\alpha} (\theta > 5^{\circ})}{N_{\alpha} (\theta > 25^{\circ})} = 28.4

A beam of 7 MeV α particles scatters from a thin gold foil. What is the ratio of the number of α particles scattered at angles greater than 5° to the number scattered at angles greater than 25°?
(Nα (θ> 5^∘))/(Nα (θ> 25^∘)) = 28.4

Added by Jorge R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

03/06/2023

Step 1

The Rutherford scattering formula is given by: dσ/dΩ = (Z1*Z2*e^2) / (4*π*ε0*E*sin^4(θ/2)) Where: dσ/dΩ = differential cross section Z1 = atomic number of incident particle (alpha particle) Z2 = atomic number of target nucleus (gold) e = elementary charge ε0 = Show more…

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I know the answer is 25.8 , show the work as to how this answer was found. Show ALL work. Thanks A beam of 7 Mev a particles scatters from a thin gold foil. What is the ratio of the number of a particles scattered at angles greater than 5 to the number scattered at angles greater than25o? Ng (8>5)=[28.4 325.8 Na(8>25o)