Question

Consider the following electromagnetic waves A, B and C: (i) The wavelength of A is \(400\,\text{nm}\).
(ii) The frequency of B is \(10^{16}\,\text{s}^{-1}\).
(iii) Wave number of C is \(10^{4}\,\text{cm}^{-1}\).
The correct order of energies is:

• A. A > B > C
• B. B > A > C
• C. B > C > A
• D. C > A > B

Hint:

For electromagnetic waves, always compare {frequencies} to decide energy order. Higher frequency $\Rightarrow$ higher photon energy.

Answer 1

The Correct Option is B

To determine the correct order of energies for the electromagnetic waves A, B, and C, we need to understand the relationship between wavelength, frequency, wave number, and energy.

The energy (\(E\)) of an electromagnetic wave is related to its frequency (\(f\)) through the equation:

\(E = h \times f\)

where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\,\text{Js}\)).

Alternatively, energy is also related to wavelength (\(\lambda\)) by:

\(E = \frac{h \times c}{\lambda}\)

where \(c\) is the speed of light (\(3 \times 10^8\,\text{m/s}\)).

Similarly, the relationship between wave number (\(\nu\)) and wavelength is:

\(ν=c​/λ\) and \(E = h \times c \times \nu\)

Let's analyze the energies of A, B, and C:

1. Wave A:
   Given wavelength, \(\lambda_A = 400\,\text{nm} = 400 \times 10^{-9}\, \text{m}\).
   Energy is calculated as:
   \(E_A = \frac{h \times c}{\lambda_A}\)
   \(E_A = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}}\)
2. Wave B:
   Given frequency, \(f_B = 10^{16}\,\text{s}^{-1}\).
   Energy is calculated as:
   \(E_B = h \times f_B\)
   \(E_B = 6.626 \times 10^{-34} \times 10^{16}\)
3. Wave C:
   Given wave number, \(\nu_C = 10^4 \,\text{cm}^{-1} = 10^6\,\text{m}^{-1}\).
   Energy is calculated as:
   \(E_C = h \times c \times \nu_C\)
   \(E_C = 6.626 \times 10^{-34} \times 3 \times 10^8 \times 10^6\)

By comparing these expressions, it's clear that frequency (as in Wave B) directly influences energy more strongly than wavelength or wave number. Hence, the order of energies is:

B (highest frequency) > A > C

This is because energy is directly proportional to frequency, and inversely proportional to wavelength (\(E \propto \frac{1}{\lambda}\)). Higher frequency indicates higher energy compared to lower wavelength and wave number.

Thus, the correct option is B > A > C.