Question

Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) ${\rm{21}}\;{\rm{cm}}$ (wavelength emitted by atomic hydrogen in interstellar space).
(b) ${\rm{1057}}\;{\rm{MHz}}$ (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) ${\rm{2}}{\rm{.7}}\;{\rm{K}}$ [temperature associated with the isotropic radiation filling all space thought to be a relic of the big-bang origin of the universe].
(d) $5890\;{{\rm{A}}^ \circ } - 5896\;{{\rm{A}}^ \circ }$ [double lines of sodium]
(e) ${\rm{14}}{\rm{.4}}\;{\rm{keV}}$ [energy of a particular transition in ${}^{57}{\rm{Fe}}$ nucleus associated with a famous high-resolution spectroscopic method (${\rm{M\ddot mossbauer}}\,{\rm{spectroscopy}}$)].

Answer 1

We know that there are basically eight types of radiation in the electromagnetic spectrum. They can be categorized as Gamma-rays, X-Rays, Ultraviolet rays, Visible rays, Infrared, Microwaves, and Radio waves. The region for these rays is specified on the basis of frequency and Wavelength. For the first question, Wavelength is given. Hence, we can directly check for Wavelength in the different regions. Then for the second question, frequency is given. Hence, we can check for the frequency range in the different regions. In the third question, the temperature is given. Hence we will use the expression of Wein’s law to find the respective Wavelength for it. In the fourth part, Wavelength is given in Armstrong. Hence we will convert into metre to check its region in the electromagnetic spectrum. In the last question, energy is given, hence we will use Planck’s law to find the frequency and then we will check for its region in the electromagnetic spectrum.

Complete step by step answer:
(a) The given wavelength is ${\rm{21}}\;{\rm{cm}}$ . We know that the range of wavelength for radio waves is ${\rm{1}}\;{\rm{mm}}$ to ${\rm{100}}\;{\rm{km}}$. Hence, we can say that ${\rm{21}}\;{\rm{cm}}$ comes at the end of a short wavelength range of radio waves.
(b) ${\rm{1057}}\;{\rm{MHz}}$can be written as ${\rm{1}}{\rm{.057}} \times {\rm{1}}{{\rm{0}}^9}\;{\rm{Hz}}$ . We know that the frequency less than $3 \times {10^{11}}\;{\rm{Hz}}$, hence the given frequency comes in the region of radio waves of electromagnetic spectrum.
(c) We will use Wein’s law to find the wavelength by using the relation between temperature and wavelength.
$\lambda T = 2.898 \times {10^{ - 3}}\;{\rm{m}} \cdot {\rm{K}}$
On rearranging the above expression and substituting ${\rm{2}}{\rm{.7}}\;{\rm{K}}$ we will get the value of wavelength.
$\begin{array}{l} \lambda = \dfrac{{2.898 \times {{10}^{ - 3}}\;{\rm{m}} \cdot {\rm{K}}}}{{2.7\;{\rm{K}}}}\\\ \lambda = 1.11 \times {10^{ - 3}}\;{\rm{m}}\\\ \lambda = 0.11\;{\rm{cm}} \end{array}$
The microwaves consist of the wavelength ranging from ${\rm{1}}\;{\rm{mm}}$ to $25\,\mu {\rm{m}}$ . In centimetre it can be represented as ${\rm{0}}{\rm{.1}}\;{\rm{cm}}$ to $25 \times {10^{ - 4}}\;{\rm{cm}}$ . Hence the radiation having temperature ${\rm{2}}{\rm{.7}}\;{\rm{K}}$ will come under the region of microwaves
The wavelength corresponds to microwaves.
(d) ${\rm{1}}\;{{\rm{A}}^ \circ }$is equals to ${10^{ - 10}}\;{\rm{m}}$. Hence the given range $5890\;{{\rm{A}}^ \circ } - 5896\;{{\rm{A}}^ \circ }$ can be written as $5.890 \times {10^{ - 10}}\;{\rm{m}} - 5896 \times {10^{ - 10}}\;{\rm{m}}$. We know that yellow light of the visible spectrum has the range $565\;{\rm{nm}} - 590\,{\rm{nm}}$ . ${\rm{1}}\;{\rm{nm}}$ is equals to ${10^{ - 9}}\,{\rm{m}}$ . Hence the range of yellow light of the visible spectrum can be expressed as$565 \times {10^{ - 9}}\;{\rm{m}} - 590 \times {10^{ - 9}}\;{\rm{m}}$. Therefore, the given range in the question for double lines of the sodium comes under the yellow light of the visible spectrum.
(e) From the Planck’s formula of transition energy, we have,
$E = h\upsilon$
Where $h$ is the Planck’s constant whose value is $6.6 \times {10^{ - 34}}\;{\rm{Js}}$ and $\upsilon$ is the frequency of radiation.
Since, it is given in the question that energy is ${\rm{14}}{\rm{.4}}\;{\rm{ke}}\;{\rm{V}}$. On Substituting the energy and Planck’s constant in the above expression, we will get

\upsilon = \dfrac{{14.4 \times \dfrac{{{{10}^3}\,eV}}{{1\,eV}} \times 1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}{{6.6 \times {{10}^{ - 34}}\;{\rm{Js}}}}\\\ \upsilon = 3.4 \times {10^{18}}\;{\rm{Hz}} \end{array}$$ The frequency range for X-Ray is ${10^{17}}\,{\rm{Hz}} - {10^{18}}\;{\rm{Hz}}$ . Hence the above frequency comes in the range of X-Ray. **Note:** We can categories the waves of the electromagnetic spectrum on the basis of their wavelength range and frequency range. To find the answer to this problem, we should have prior knowledge about the ranges and regions of different waves.