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Question: A quantum of light having energy E has wavelength equal to \(7200\mathop A\limits^o \). The frequenc...

A quantum of light having energy E has wavelength equal to 7200Ao7200\mathop A\limits^o . The frequency of light which corresponds to energy equal to 3E is:
A) 1.25×1014s−1 B) 1.25×1015s−1 C) 1.25×1013s−1 D) 1.25×1016s−1  A){\text{ }}1.25 \times {10^{14}}{s^{ - 1}} \\\ B){\text{ }}1.25 \times {10^{15}}{s^{ - 1}} \\\ C){\text{ }}1.25 \times {10^{13}}{s^{ - 1}} \\\ D){\text{ }}1.25 \times {10^{16}}{s^{ - 1}} \\\

Explanation

Solution

The wavelength is defined as the distance which is present between the consecutive crest of a wave. We can calculate wavelength by dividing velocity by frequency. The shorter the wavelengths, the higher the frequency and it indicates greater energy. So the longer the wavelengths and lower the frequency results in lower energy.

Complete step by step answer:
Let us understand the terms in the question first.
A quantum of light is called a photon. Photons are minute energy packets of electromagnetic waves which are also called light quantum.
Wavelength is known as the distance between two successive crests or troughs of a wave. Wavelength is inversely proportional to frequency.
The frequency is the number of waves that pass a certain point in a specified amount of time.
Now let us understand the relation between wavelength and frequency.
Wavelength is inversely proportional to frequency. So according to this longer wavelength, lower the frequency. So we can also say that shorter the wavelength, higher will be the frequency.
As we know that the wavelength and frequency are related to light, we can also relate them to energy. The shorter the wavelengths, the higher the frequency and it indicates greater energy. So the longer the wavelengths and lower the frequency results in lower energy.
The energy of the equation can be written as E = hν.
According to the question
A quantum of light having energy E has wavelength equal to 7200Ao7200\mathop A\limits^o . Also, energy equal to 3E.

wavelength=13(7200Ao)wavelength = \dfrac{1}{3}(7200{A^o})
ν=cλ=3×3×108m/s7200×10−10m\nu = \dfrac{c}{\lambda } = \dfrac{{3 \times 3 \times {{10}^8}m/s}}{{7200 \times {{10}^{ - 10}}m}}
=1.25×1015s−1= 1.25 \times {10^{15}}{s^{ - 1}}

Note: The frequency is defined as the time period or the time which is taken to complete the cycle of the wave to a point. It is the number of cycles which have been completed in a unit time. The angular frequency is the angular displacement of the element of the wave present in per unit time.