Compare the energies of two radiations `E_(1)` with wavelength 800 nm and `E_(2)` with wavelength 400 nm.

Options

A. `E_(1)=2E_(2)`
B. `E_(1)=E_(2)`
C. `E_(2)=2E_(1)`
D. `E_(2)=-1/2E_(1)`

Explanation

To compare the energies of two radiations \( E_1 \) with a wavelength of 800 nm and \( E_2 \) with a wavelength of 400 nm, we can use the formula for energy of a photon:

\[ E = \frac{hc}{\lambda} \]

where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.

### Step 1: Identify the wavelengths
- For \( E_1 \): \( \lambda_1 = 800 \, \text{nm} = 800 \times 10^{-9} \, \text{m} \)
- For \( E_2 \): \( \lambda_2 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)

### Step 2: Write the energy equations
Using the formula \( E = \frac{hc}{\lambda} \):
- For \( E_1 \):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{hc}{800 \times 10^{-9}} \]
- For \( E_2 \):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{hc}{400 \times 10^{-9}} \]

### Step 3: Compare the energies
Since \( hc \) is a constant, we can compare the energies directly by looking at the wavelengths:
\[ \frac{E_2}{E_1} = \frac{\frac{hc}{\lambda_2}}{\frac{hc}{\lambda_1}} = \frac{\lambda_1}{\lambda_2} \]

Substituting the values:
\[ \frac{E_2}{E_1} = \frac{800 \times 10^{-9}}{400 \times 10^{-9}} = 2 \]

This means:
\[ E_2 = 2 E_1 \]

### Conclusion
The energy of radiation \( E_2 \) (400 nm) is twice that of radiation \( E_1 \) (800 nm):
\[ E_2 = 2 E_1 \]