Question

Consider the following electromagnetic waves: wave A :- wavelength = \(400\,\text{nm}\)
wave B :- frequency = \(10^{16}\,\text{Hz}\)
wave C :- wave number = \(10^{4}\,\text{cm}^{-1}\)
order of energy is :

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

Hint:

For electromagnetic waves, compare {frequencies directly} to decide energy order — higher frequency always means higher photon energy.

Answer 1

The Correct Option is C

To determine the order of energy for the given electromagnetic waves, we begin by recalling that the energy of a wave is directly proportional to its frequency and inversely proportional to its wavelength. The energy \( E \) of an electromagnetic wave can be described by the equation:

E = h \nu = \frac{hc}{\lambda},

where:

• E is the energy of the wave,
• h is Planck's constant,
• \nu is the frequency of the wave,
• \lambda is the wavelength of the wave,
• c is the speed of light in a vacuum.

Let us calculate the energy associated with each type of electromagnetic wave:

1. Wave A: Given wavelength \lambda_A = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}. Using E_A = \frac{hc}{\lambda_A}, we find a high energy since wavelength is inversely proportional to energy.
2. Wave B: Given frequency \nu_B = 10^{16} \, \text{Hz}. Using E_B = h\nu_B, this wave has a very high energy due to its high frequency directly proportional to its energy.
3. Wave C: Given wave number k_C = 10^{4} \, \text{cm}^{-1} = 10^{6} \, \text{m}^{-1}. Recall the relation k = \frac{1}{\lambda}, so \lambda_C = \frac{1}{k_C} = 10^{-6} \, \text{m}. Similar to wave A, use E_C = \frac{hc}{\lambda_C}, resulting in lower energy than waves A and B.

By comparing the above energies:

• Wave B has the highest energy because it has the highest frequency.
• Wave A has the next highest energy due to its lesser wavelength compared to wave C.
• Wave C has the least energy based on both its wave number and resulting wavelength.

Therefore, the energy order is B > A > C.