The alpha particles emitted by radium have energies of 4.795 and 4.611 MeV. What is the wavelength o — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

The energy of the gamma ray is $\Delta E = 4.795 \text{ MeV} - 4.611 \text{ MeV} = 0.184 \text{ MeV} = 0.184 \times 10^6 \text{ eV}$. Using the relation $\lambda = \frac{hc}{\Delta E}$, where $hc \approx 1240 \text{ eV}\cdot\text{nm}$: $\lambda = \frac{1240}{0.184 \times 10^6} \text{ nm} \approx 6.739 \times 10^{-3} \text{ nm}$. Converting to picometers, $\lambda \approx 6.74 \text{ pm}$. Therefore, correct answer is A.