Question 1[Maximum number: 3]Question (a)(a)Show that the energy of the scattered photon is about 0.17 MeV .[ 1 ]Show AnswerE=<hcλ=>1.24×10−67.47×10−12E=<\frac{h c}{\lambda}=>\frac{1.24 \times 10^{-6}}{7.47 \times 10^{-12}}E=<λhc=>7.47×10−121.24×10−6 OR 0.166 MeVQuestion (b)(b)For an electron, hmec=2.43×10−12 m\frac{h}{m_{\mathrm{e}} c}=2.43 \times 10^{-12} \mathrm{~m}mech=2.43×10−12 m. Determine θ\thetaθ.[ 2 ]Show Answer7.47×10−12−6.40×10−12=2.43×10−12(1−cosθ)7.47 \times 10^{-12}-6.40 \times 10^{-12}=2.43 \times 10^{-12}(1-\cos \theta)^{}7.47×10−12−6.40×10−12=2.43×10−12(1−cosθ)« cosθ=0.5597\cos \theta=0.5597cosθ=0.5597 so » θ=56.0o‾\theta=56.0^{\underline{\mathrm{o}}}θ=56.0oAccept angle in radians ( 0.98 « rad »).Award [2] if 56 « ∘{ }^{\circ}∘ » or 0.98 « rad» are seen as the answer without workingAdd to Test
Question (a)(a)Show that the energy of the scattered photon is about 0.17 MeV .[ 1 ]Show AnswerE=<hcλ=>1.24×10−67.47×10−12E=<\frac{h c}{\lambda}=>\frac{1.24 \times 10^{-6}}{7.47 \times 10^{-12}}E=<λhc=>7.47×10−121.24×10−6 OR 0.166 MeV
Question (b)(b)For an electron, hmec=2.43×10−12 m\frac{h}{m_{\mathrm{e}} c}=2.43 \times 10^{-12} \mathrm{~m}mech=2.43×10−12 m. Determine θ\thetaθ.[ 2 ]Show Answer7.47×10−12−6.40×10−12=2.43×10−12(1−cosθ)7.47 \times 10^{-12}-6.40 \times 10^{-12}=2.43 \times 10^{-12}(1-\cos \theta)^{}7.47×10−12−6.40×10−12=2.43×10−12(1−cosθ)« cosθ=0.5597\cos \theta=0.5597cosθ=0.5597 so » θ=56.0o‾\theta=56.0^{\underline{\mathrm{o}}}θ=56.0oAccept angle in radians ( 0.98 « rad »).Award [2] if 56 « ∘{ }^{\circ}∘ » or 0.98 « rad» are seen as the answer without working