Question 23M.2.HL.TZ1.8

Use of E_max = $\frac{hc}{\lambda }-\varphi \Rightarrow \varphi =\frac{hc}{\lambda }-$ E_max ✓

$\varphi =$«$\frac{hc}{\lambda }$ − E_max = $\frac{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(468\times {10}^{-9}\right)\left(1.6\times {10}^{-19}\right)}$ − 1.81» = 0.85625 $\approx$ 0.86 «eV» ✓

Use of $\frac{hc}{\lambda }=\varphi \Rightarrow \lambda =\frac{hc}{\varphi }$ ✓

$\lambda =$«$\frac{\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(468\times {10}^{-9}\right)\left(1.6\times {10}^{-19}\right)}$ =» 1.45 × 10⁻⁶ «m»✓

Allow ECF from a(i)

2e AND 82e seen

OR

3.2 × 10⁻¹⁹ «C» AND 1.312 × 10⁻¹⁷ «C» seen ✓

d = $\frac{8.99\times {10}^9\times \left(2e\right)\left(82e\right)}{5.9\times {10}^6\times e}$ = 3.998 × 10⁻¹⁴ ≈ 4 × 10⁻¹⁴ «m» ✓

Must see either clear substitutions or answer to at least 4 s.f. for MP2.

The closest approach is «significantly» larger than the radius of the nucleus / far away from the nucleus/OWTTE. ✓

«Therefore» the strong nuclear force will not act on the alpha particle.✓