Compare the energies of two radiation one with a wavelength of `300 nm` and other with `600 nm`.

Answer

To compare the energies of two radiations with wavelengths of 300 nm and 600 nm, we can use the formula for energy associated with electromagnetic radiation:

### Step-by-Step Solution:

1. **Understand the Energy Formula**:
The energy \( E \) of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) = energy of the photon
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) = speed of light (\( 3.00 \times 10^8 \, \text{m/s} \))
- \( \lambda \) = wavelength of the radiation

2. **Calculate Energy for Each Wavelength**:
- For the first radiation with wavelength \( \lambda_1 = 300 \, \text{nm} \):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{hc}{300 \times 10^{-9} \, \text{m}} \]
- For the second radiation with wavelength \( \lambda_2 = 600 \, \text{nm} \):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{hc}{600 \times 10^{-9} \, \text{m}} \]

3. **Set Up the Ratio of Energies**:
To compare the energies, we can take the ratio \( \frac{E_1}{E_2} \):
\[ \frac{E_1}{E_2} = \frac{\frac{hc}{300 \times 10^{-9}}}{\frac{hc}{600 \times 10^{-9}}} \]

4. **Simplify the Ratio**:
The \( hc \) terms will cancel out:
\[ \frac{E_1}{E_2} = \frac{600 \times 10^{-9}}{300 \times 10^{-9}} = \frac{600}{300} = 2 \]

5. **Conclusion**:
This means that the energy of the radiation with a wavelength of 300 nm is twice that of the radiation with a wavelength of 600 nm. Therefore:
\[ E_1 = 2 \times E_2 \]

### Final Answer:
The energy of the radiation with a wavelength of 300 nm is twice the energy of the radiation with a wavelength of 600 nm.

---