A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 4990 A and 5000A is `E_(1)` , and that between 9990 A and 10000 A is `E_(2)` . Find the ratio of `E_(2)` and `E_(1)`.

Explanation

To find the ratio of the energies \( E_2 \) and \( E_1 \) emitted by a black body at a temperature of 2880 K for the specified wavelengths, we will use Planck's Radiation Law.

### Step-by-Step Solution:

1. **Understand Planck's Radiation Law**:
The energy emitted per unit wavelength by a black body is given by:
\[
E_{\lambda} = \frac{2 \pi h c^2}{\lambda^5} \cdot \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}
...