Question

An atom is initially at an energy level $$E=-6.52eV$$. It absorbs a photon of wavelength $$860nm$$. The internal energy of atom after absorbing photon is
A) 5.08eV
B) 1.44eV
C) -1.44eV
D) -5-08eV

Answer 1

Hint: In this question, we use the concept of photon energy, where the energy carried by a single photon is the product of frequency by Planck’s constant. After finding the energy of the photon, we will find the energy resultant after the atom absorbs a photon by subtracting the atom’s energy level or binding energy by the energy of the photon. The formula for the energy of photon is:
$$E_p=hf$$
The energy of the atom $$\left(E_a\right)$$ after absorbing the photon energy $$\left(E_{abs}\right)$$ is:
$$E_{abs}=E_p+E_a$$
where $$E_p$$ is the energy of the photon received, $$h$$ is the Planck’s constant of $$6.626\times {10}^{-34}j$$, $$f$$ is the frequency that can also be written as $${\displaystyle \frac{c}{\lambda }}$$, $$c$$ is the speed of light at $$288000m/\sec$$, wavelength of a photon absorbed $$860nm$$.

Complete step by step solution:
The energy carried by the photon is inversely proportional to the wavelength of the photon absorbed; the formula for the energy is given as:
$$E_p=hf$$
Now, the frequency of the energy of the photon is
$$f={\displaystyle \frac{c}{\lambda }}$$
Placing the value of the $$\lambda$$ and the speed of light, we get the value of the frequency as:
$$f={\displaystyle \frac{c}{\lambda }}$$
$$f={\displaystyle \frac{288000}{860}}$$
Now multiplying the value of the frequency by the Planck’s constant, we get the energy of the photon as:
$$E_p=6.626\times {10}^{-34}\times {\displaystyle \frac{288000}{860}}$$
$$=1.44eV$$
The energy generated by the photon when moving through the energy levels is given as $$1.44eV$$.
Now as we know the energy of the photon, we can find the energy of the atom once the photon is absorbed by the atom. Hence, the net energy after the absorption of the photon is given as:
The net energy of the atom is equal to the subtraction of energy of the photon by the previous energy of the atom.
$$E_{abs}=E_p-E_a$$
$$E_{abs}=1.44+(-6.52)$$
$$E_{abs}=-5.08eV$$
Therefore, the energy formed after absorbing the photon is $$E_{abs}=-5.08eV$$.

Note: The energy of the atom is also known as the binding energy, the energy that the atom has in the level that is in before absorbing or emission of energy. When the atom absorbs a photon it either jumps to another level or recedes to another level. Now, another formula to find the energy of the photon is based on the number of energy levels i.e.
$$E=-{\displaystyle \frac{13.6}{n}}eV$$
But here we use the formula:
$$E_p=hf$$
To find the energy level as the number of levels or rings of the atom are not known.