Question 1[Maximum number: 5]Question (a)(a)Electrons are accelerated through a potential difference of 15 kV . The electrons collide with a metal target and a spectrum of X-rays is produced.[ 5 ]Question (i)(i)Explain why a continuous spectrum of energies of X-ray photons is produced.[ 3 ]Mark as masteredShow Answerelectrons decelerate (on hitting target) so X-ray photons producedrange of decelerationsphoton energy depends on (magnitude of) decelerationQuestion (ii)(ii)Calculate the wavelength of the highest energy X-ray photon produced.[ 2 ]Show AnswereV=hcλe V=\frac{h c}{\lambda}eV=λhcλ=6.63×10−34×3.0×1081.6×10−19×15000\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6 \times 10^{-19} \times 15000}λ=1.6×10−19×150006.63×10−34×3.0×108=8.3×10−11 m=8.3 \times 10^{-11} \mathrm{~m}=8.3×10−11 morE=h f and c=fλc=f \lambdac=fλ and electron energy =eVorE=hc/λE=h c / \lambdaE=hc/λ and electron energy =eV electron energy =1.6×10−19×15000=2.4×10−15\begin{aligned} \text { electron energy } & =1.6 \times 10^{-19} \times 15000 & =2.4 \times 10^{-15} \end{aligned} electron energy =1.6×10−19×15000=2.4×10−15λ=6.63×10−34×3.0×1082.4×10−15\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{-15}}λ=2.4×10−156.63×10−34×3.0×108λ=8.3×10−11 m\lambda=8.3 \times 10^{-11} \mathrm{~m}λ=8.3×10−11 mAdd to Test
Question (a)(a)Electrons are accelerated through a potential difference of 15 kV . The electrons collide with a metal target and a spectrum of X-rays is produced.[ 5 ]Question (i)(i)Explain why a continuous spectrum of energies of X-ray photons is produced.[ 3 ]Mark as masteredShow Answerelectrons decelerate (on hitting target) so X-ray photons producedrange of decelerationsphoton energy depends on (magnitude of) decelerationQuestion (ii)(ii)Calculate the wavelength of the highest energy X-ray photon produced.[ 2 ]Show AnswereV=hcλe V=\frac{h c}{\lambda}eV=λhcλ=6.63×10−34×3.0×1081.6×10−19×15000\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6 \times 10^{-19} \times 15000}λ=1.6×10−19×150006.63×10−34×3.0×108=8.3×10−11 m=8.3 \times 10^{-11} \mathrm{~m}=8.3×10−11 morE=h f and c=fλc=f \lambdac=fλ and electron energy =eVorE=hc/λE=h c / \lambdaE=hc/λ and electron energy =eV electron energy =1.6×10−19×15000=2.4×10−15\begin{aligned} \text { electron energy } & =1.6 \times 10^{-19} \times 15000 & =2.4 \times 10^{-15} \end{aligned} electron energy =1.6×10−19×15000=2.4×10−15λ=6.63×10−34×3.0×1082.4×10−15\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{-15}}λ=2.4×10−156.63×10−34×3.0×108λ=8.3×10−11 m\lambda=8.3 \times 10^{-11} \mathrm{~m}λ=8.3×10−11 m
Question (i)(i)Explain why a continuous spectrum of energies of X-ray photons is produced.[ 3 ]Mark as masteredShow Answerelectrons decelerate (on hitting target) so X-ray photons producedrange of decelerationsphoton energy depends on (magnitude of) deceleration
Question (ii)(ii)Calculate the wavelength of the highest energy X-ray photon produced.[ 2 ]Show AnswereV=hcλe V=\frac{h c}{\lambda}eV=λhcλ=6.63×10−34×3.0×1081.6×10−19×15000\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6 \times 10^{-19} \times 15000}λ=1.6×10−19×150006.63×10−34×3.0×108=8.3×10−11 m=8.3 \times 10^{-11} \mathrm{~m}=8.3×10−11 morE=h f and c=fλc=f \lambdac=fλ and electron energy =eVorE=hc/λE=h c / \lambdaE=hc/λ and electron energy =eV electron energy =1.6×10−19×15000=2.4×10−15\begin{aligned} \text { electron energy } & =1.6 \times 10^{-19} \times 15000 & =2.4 \times 10^{-15} \end{aligned} electron energy =1.6×10−19×15000=2.4×10−15λ=6.63×10−34×3.0×1082.4×10−15\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{-15}}λ=2.4×10−156.63×10−34×3.0×108λ=8.3×10−11 m\lambda=8.3 \times 10^{-11} \mathrm{~m}λ=8.3×10−11 m