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Study E.3.17 (HL)—Discrete nuclear energy levels conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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Rhodium-106 $\left({ }_{45}^{106} \mathrm{Rh}\right)$ decays into palladium-106 $\left({ }_{46}^{106} \mathrm{Pd}\right)$ by beta minus $\left(\beta^{-}\right)$ decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the $\beta^{-}$ decay.

Calculate the wavelength of the gamma ray photon in (d)(ii).

Photon energy

$\begin{aligned} E&=0.48\times10^6\times1.6\times10^{-19}=7.68\times10^{-14}\ \mathrm{J}\\ \lambda&=\frac{hc}{E}=\frac{6.63\times10^{-34}\times3\times10^8}{7.68\times10^{-14}}=2.6\times10^{-12}\ \mathrm{m} \end{aligned}$

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IB Physics HL E 3 17 Hl Discrete Nuclear Energy Levels Topic Practice

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