An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times {10^{ - 2}}{\rm{A}}.$ The maximum energy of a photon in the radiation in eV is $h = 6.62 \times {10^{ - 34}}$ Js , $c = 3 \times {10^8}\;{\rm{m}}{{\rm{s}}^{ - 1}})$

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Energy of a photon is inversely proportional to its wavelength.\n\t**Related Topics**: Electromagnetic spectrum, Planck's quantum theory.\n\t**Prerequisites**: Understanding of electromagnetic radiation, relationship between energy, frequency, and wavelength, and unit conversion.\n## SOLUTION PATHWAY\n**Given Information**:\n\tShortest wavelength (λ) = $45 \times 10^{-2} Å = 45 \times 10^{-2} \times 10^{-10} m = 45 \times 10^{-12} m$\n\tPlanck's constant (h) = $6.62 \times 10^{-34} Js$\n\tSpeed of light (c) = $3 \times 10^8 m/s$\n\t\n**Method & Steps**:\n\t1. Calculate the energy of the photon in Joules using the formula $E = \frac{hc}{\lambda}$:\n\t\tReasoning: The maximum energy corresponds to the shortest wavelength.\n\t\tFormula/Rule: $E = \frac{hc}{\lambda}$\n\t2. Substitute the given values:\n\t\tWorking: $E = \frac{(6.62 \times 10^{-34} Js)(3 \times 10^8 m/s)}{45 \times 10^{-12} m} = \frac{19.86 \times 10^{-26}}{45 \times 10^{-12}} J = 0.4413 \times 10^{-14} J$\n\t\tKey Point: Ensure consistent units (meters for wavelength)\n\t3. Convert the energy from Joules to electron volts (eV) using the conversion factor $1 eV = 1.6 \times 10^{-19} J$:\n\t\tWorking: $E(eV) = \frac{0.4413 \times 10^{-14} J}{1.6 \times 10^{-19} J/eV} = 0.2758 \times 10^5 eV = 27580 eV$\n\t\tResult: $E \approx 27500 eV$\n\t\tVerification: The calculated value is close to option A\n## OPTION ANALYSIS\n[A] ✓: The calculated maximum energy is approximately 27500 eV, matching the correct answer.\n[B] :x: The calculated value is not 22500 eV.\n[C] :x: The calculated value is not 17500 eV.\n[D] :x: The calculated value is not 12500 eV.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Incorrect unit conversion of wavelength from Angstrom to meters. How to avoid: Always convert all units into SI units before calculation.\n\t\t2.[Error 2]: Forgetting to convert the energy from joules to eV. How to avoid: Remember the conversion factor $1 eV = 1.6 \times 10^{-19} J$.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Memorize the relation that $E = \frac{1240}{\lambda(nm)} eV$. Convert wavelength to nm and plug it into the formula.\n\t\t2.[Memory aid]: Remember the energy of a photon is inversely related to the wavelength.

An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times {10^{ - 2}}{\rm{A}}.$ The maximum energy of a photon in the radiation in eV is $h = 6.62 \times {10^{ - 34}}$ Js , $c = 3 \times {10^8}\;{\rm{m}}{{\rm{s}}^{ - 1}})$

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Relationship between photon energy and wavelength in X-ray production.\n\t**Related Topics**: Electromagnetic spectrum, Planck's equation, energy conversion.\n\t**Prerequisites**: Understanding of X-ray production, the relationship between energy and wavelength of photons, and the concept of electron volt (eV).\n## SOLUTION PATHWAY\n**Given Information**:\n\tShortest wavelength, $\lambda = 45 \times 10^{-12} m$ (Note: $1 \text{A} = 10^{-10} \text{m}$ so $45 \times 10^{-2} \text{A} = 45 \times 10^{-2} \times 10^{-10} \text{m} = 45 \times 10^{-12} \text{m}$)\n\tPlanck's constant, $h = 6.62 \times 10^{-34} \text{Js}$\n\tSpeed of light, $c = 3 \times 10^8 \text{m/s}$\n**Method & Steps**:\n\t1. Calculate the energy of the photon in Joules using the formula $E = \frac{hc}{\lambda}$.\n\t\tReasoning: The shortest wavelength corresponds to the highest energy photon.\n\t\tFormula/Rule: $E = \frac{hc}{\lambda}$\n\t2. Convert the energy from Joules to electron volts (eV).\n\t\tWorking: $E = \frac{(6.62 \times 10^{-34} \text{Js}) \times (3 \times 10^8 \text{m/s})}{45 \times 10^{-12} \text{m}} = 4.4133 \times 10^{-14} \text{J}$ \n\t\tKey Point: $1 \text{eV} = 1.602 \times 10^{-19} \text{J}$, so we divide by this value to get energy in eV\n\t3. Calculate the energy in eV: $E_{eV} = \frac{4.4133 \times 10^{-14} \text{J}}{1.602 \times 10^{-19} \text{J/eV}} = 27550 \text{eV} \approx 27500 \text{eV}$\n\t\tResult: The maximum energy of the photon is approximately 27500 eV.\n\t\tVerification: The calculated value matches the correct answer.\n## OPTION ANALYSIS\n[A] ✓: The calculated value closely matches the correct answer of 27500 eV.\n[B] :x: The value 22500 eV is significantly lower than the calculated result.\n[C] :x: The value 17500 eV is much lower than the calculated result.\n[D] :x: The value 12500 eV is much lower than the calculated result.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Using the wrong wavelength value : Ensure that you convert from Angstrom to meters.\n\t\t2. Forgetting to convert from Joules to electron volts: Always verify the units requested in the answer.\n\t**Quick Tips**:\n\t\t1. Remember the formula $E=hc/\lambda$ to calculate photon energy.\n\t\t2. Memorize the conversion factor between Joules and electron volts, $1 \text{eV} = 1.602 \times 10^{-19} \text{J}$.