Question 705073

SCQHARD

In a scattering experiment a beam of e−e^- with an energy of 420 MeV scatters off an atomic nucleus. If the first minimum of the differential cross section is observed at a scattering angle of 45∘45^\circ, the radius of the nucleus (in fermi) is closest to

(A)

0.4

(B)

8.0

(C)

2.5

(D)

0.8

Detailed Solution

Wavelength λ=hcE=1240 MeV fm420 MeV≈2.95 fm\lambda = \frac{hc}{E} = \frac{1240 \text{ MeV fm}}{420 \text{ MeV}} \approx 2.95 \text{ fm}. Condition for first minimum: sin⁡θ=1.22λd\sin \theta = \frac{1.22\lambda}{d}. With θ=45∘\theta = 45^\circ, d=1.22×2.95sin⁡45∘≈5.09 fmd = \frac{1.22 \times 2.95}{\sin 45^\circ} \approx 5.09 \text{ fm}. Radius r=d2≈2.54 fmr = \frac{d}{2} \approx 2.54 \text{ fm}.

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