Question

What is the frequency of a photon with a wavelength of $2.5\text{ x 1}{{\text{0}}^{-9}}m$? In what region of the spectrum is this located?

Answer 1

Hint: We can use the formula to solve this question is:
$c=\lambda v$
Where c is the speed of light, $\lambda $ is the wavelength of the photon and $v$ is the frequency of the photon.

Complete answer:
We know that the radiant energy emitted or absorbed in the form of small packets is known as photon and we also know that the photon travels with a speed of light which is equal to $3\text{ x 1}{{\text{0}}^{8}}\text{ m/s}$. So, we can use the formula to solve this question is:
$c=\lambda v$
Where c is the speed of light, $\lambda $ is the wavelength of the photon and $v$ is the frequency of the photon.
Now, we are given the value of wavelength as $2.5\text{ x 1}{{\text{0}}^{-9}}m$ and the value of speed of light is known. So, when we put the value in the formula we get the value of frequency:
 $3\text{ x 1}{{\text{0}}^{8}}=2.5\text{ x 1}{{\text{0}}^{-9}}\text{ x }v$
$v=\dfrac{3\text{ x 1}{{\text{0}}^{8}}}{2.5\text{ x 1}{{\text{0}}^{-9}}}=1.2\text{ x 1}{{\text{0}}^{17}}Hz$
So, the frequency of the photon is $1.2\text{ x 1}{{\text{0}}^{17}}$ Hertz.
As we can see that the wavelength of the photon is small and the frequency of the photon is very high, so it lies in the soft X-rays region of the electromagnetic spectrum. Since the frequency of the photon is high, the energy of the photon will be high because the energy of the photon is directly proportional to the frequency of the photon.

Note:
The radiation that has the highest energy or highest frequency or lowest wavelength are cosmic rays, while the radiation that has the lowest energy or the lowest frequency or highest wavelength are radio waves.