The energy difference between the ground state of an atom and its excited state is $3 \times {10^{ - 19}}{\text{ J}}$. What is the wavelength of the photon required for this transition?
A. $494.5 \times {10^{ - 6}}{\text{ m}}$
B. $3 \times {10^{ - 8}}{\text{ m}}$
C. $1.8 \times {10^{ - 7}}{\text{ m}}$
D. $6.6 \times {10^{ - 7}}{\text{ m}}$

Hint:The energy of the photon is inversely proportional to the wavelength and thus, directly proportional to the frequency. This is because wavelength and frequency are inversely proportional to each other. As the wavelength increases, the energy of the photon decreases.

Formula Used: $E = \dfrac{{hc}}{\lambda }$

Complete answer:
We know that the relation between energy and wavelength of a photon is given by the Planck-Einstein equation as follows:
$E = \dfrac{{hc}}{\lambda }$
Where,
$E$ is the energy of the photon,
$h$ is the Planck’s constant,
$c$ is the velocity of light in vacuum,
$\lambda $ is the wavelength of the photon.
Rearrange the equation for the wavelength of the photon as follows:
$\lambda = \dfrac{{hc}}{E}$
Substitute $6.626 \times {10^{ - 34}}{\text{ J s}}$ for the Planck’s constant, $3 \times {10^8}{\text{ m }}{{\text{s}}^{ - 1}}$ for the velocity of light in vacuum, $3 \times {10^{ - 19}}{\text{ J}}$ for the energy of the photon and solve for the wavelength of the photon. Thus,
$\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}{\text{ J s}} \times 3 \times {{10}^8}{\text{ m }}{{\text{s}}^{ - 1}}}}{{3 \times {{10}^{ - 19}}{\text{ J}}}}$
$\Rightarrow \lambda = 6.626 \times {10^{ - 7}}{\text{ m}}$
Thus, the wavelength of the photon required for this transition is $6.626 \times {10^{ - 7}}{\text{ m}}$.

Thus, the correct option is (D) $6.6 \times {10^{ - 7}}{\text{ m}}$.

Note:
Tiny particles having no charge and no resting mass are known as photons. The photons are emitted by charged particles, radioactive decay, etc. photons always move at the speed of light in vacuum. Photons can be destroyed as well as created. When electromagnetic waves are emitted by the source, photons are created. When photons hit with matter, they either absorb or transfer the energy to the atoms and molecules. The creation and destruction of photons conserves energy and momentum.