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Step-by-Step Explanation
Step 1: Recall the Energy-Wavelength Relation
The energy of any photon is given by the relation
$E = \frac{hc}{\lambda}$ ,
where
• $E$ is the photon’s energy,
• $h$ is Planck’s constant,
• $c$ is the speed of light, and
• $\lambda$ is the wavelength of the emitted photon.
Step 2: Write Down the Energies for Nuclear and Atomic Transitions
• Let $E_N$ be the energy of the photon emitted by the nucleus when it de-excites. Its wavelength is $\lambda_N$.
• Let $E_A$ be the energy of the photon emitted by the atom when it de-excites. Its wavelength is $\lambda_A$.
Using the formula,
$E_N = \frac{hc}{\lambda_N}$ and $E_A = \frac{hc}{\lambda_A}.$
Step 3: Form the Ratio of Wavelengths
We want the ratio $\frac{\lambda_N}{\lambda_A}$. From the energy expressions:
\[
\frac{E_A}{E_N}
= \frac{\frac{hc}{\lambda_A}}{\frac{hc}{\lambda_N}}
= \frac{\lambda_N}{\lambda_A}.
\]
So,
\[
\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N}.
\]
Step 4: Compare Typical Energy Ranges for Atomic vs Nuclear Transitions
• Atomic transitions typically have energies of the order of electronvolts (eV).
• Nuclear transitions typically have energies of the order of mega-electronvolts (MeV), i.e. $1\,\mathrm{MeV} = 10^6\,\mathrm{eV}$.
Step 5: Determine the Numerical Ratio
Since $E_A$ is in eV and $E_N$ is in MeV,
\[
\frac{E_A}{E_N}
= \frac{\text{(some eV value)}}{\text{(some MeV value)}}
= \frac{1\,\mathrm{eV}}{10^6\,\mathrm{eV}}
= 10^{-6}.
\]
Therefore,
\[
\frac{\lambda_N}{\lambda_A} = 10^{-6}.
\]
Final Answer
The closest ratio $\frac{\lambda_{N}}{\lambda_{A}}$ is $10^{-6}$.
