Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths `lambda_(N) ,lambda_(A)` respectively. The ratio `(lambda_(N))/(lambda_(A))` is closest to :

Options

A. `10^(-6)`
B. `10^(-1)`
C. `10^(-10)`
D. 10

Answer

To solve the problem, we need to find the ratio of the wavelengths of the photons emitted during the de-excitation of the nucleus and the atom, denoted as \( \frac{\lambda_N}{\lambda_A} \).

### Step-by-Step Solution:

1. **Understanding the Relationship Between Energy and Wavelength**:
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. From this relationship, we can see that energy is inversely proportional to wavelength:
\[ \lambda \propto \frac{1}{E} \]

2. **Setting Up the Ratio**:
To find the ratio of the wavelengths, we can express it in terms of energy:
\[ \frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N} \]

3. **Substituting the Energy Values**:
We are given the energies of the excited states:
- The energy of the atom \( E_A = 1 \, \text{eV} \)
- The energy of the nucleus \( E_N = 1 \, \text{MeV} = 10^6 \, \text{eV} \)

Now substituting these values into the ratio:
\[ \frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N} = \frac{1 \, \text{eV}}{1 \, \text{MeV}} = \frac{1 \, \text{eV}}{10^6 \, \text{eV}} = 10^{-6} \]

4. **Final Result**:
Thus, the ratio of the wavelengths is:
\[ \frac{\lambda_N}{\lambda_A} = 10^{-6} \]

### Conclusion:
The closest value for the ratio \( \frac{\lambda_N}{\lambda_A} \) is \( 10^{-6} \).

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