To solve this problem, we need to understand the relationship between the energy of a photon and its wavelength. The energy of a photon (E) can be calculated using the equation:

$$E = \frac{hc}{\lambda}$$

where:

• $E$ is the energy of the photon,


• $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J s}$),


• $c$ is the speed of light in vacuum ($3.0 \times 10^8 \, \text{m/s}$),


• $\lambda$ is the wavelength of the photon.

Let's denote the energy of a photon with a wavelength of 2000Å as $E_1$, and the energy of a photon with a wavelength of 4000Å as $E_2$. We need to find the ratio $\frac{E_1}{E_2}$.

First, convert wavelengths from Ångströms to meters, since the constants are in SI units:

1Å = $10^{-10}$ meters.

• So, 2000Å = $2000 \times 10^{-10}$ meters = $2 \times 10^{-7}$ meters.


• And 4000Å = $4000 \times 10^{-10}$ meters = $4 \times 10^{-7}$ meters.

Next, we can substitute the values into the energy equation for both $E_1$ and $E_2$ and then find their ratio:

$$E_1 = \frac{hc}{\lambda_1} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2 \times 10^{-7}}$$

$$E_2 = \frac{hc}{\lambda_2} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{4 \times 10^{-7}}$$

Now, let's find the ratio $\frac{E_1}{E_2}$:

$$\frac{E_1}{E_2} = \frac{\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2 \times 10^{-7}}}{\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{4 \times 10^{-7}}} = \frac{4 \times 10^{-7}}{2 \times 10^{-7}} = 2$$

Therefore, the ratio of the energy of a photon of 2000Å wavelength to that of a 4000Å radiation is 2. Hence, the correct answer is:

Option D: 2