Question

The spectrum of radiation \[1.0 \times {10^{14}}\;Hz\] is the infrared region. The energy of one photon of this in joules will be
A. \[6.62 \times {10^{ - 48}}\]
B. \[6.62 \times {10^{ - 20}}\]
C. \[\dfrac{{6.62}}{3} \times {10^{ - 28}}\]
D. \[3 \times 6.62 \times {10^{ - 28}}\]

Answer 1

Hint: Planck was the first to give the formula for quantisation of energy. The dual nature (particle and wave) of light means that sometimes it behaves as a particle and sometimes as a wave. Planck stated that light consists of quanta (or particles) called photons with each such quanta having energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]. Phenomenon such as the photoelectric effect also prove that light behaves as particles as well. Phenomenon such as refraction and transmission, the behaviour of light is that of a wave.

Formula Used:
Energy, \[E = h\nu \]
\[\nu = \dfrac{c}{\lambda }\]
Where,
E = Energy of the photon
\[c{\rm{ }}\] = speed of light = \[3 \times {10^8}m/s\]
\[\nu \]= frequency of the light (in Hz)
\[\lambda \]= wavelength of the light (in m)

Complete step by step solution:
Based on the frequency, the light spectrum is divided into three parts, ultraviolet (UV), Visible and Infrared (IR) regions in order of decreasing frequency.. We can only see the light radiation using our eyes, whose frequency lies in the visible region. The light spectrum is also called the electromagnetic spectrum.

Given: A spectrum of radiation of wavelength equal to \[1.0 \times {10^{14}}\;Hz\] is the infrared region. We need to determine the energy of photons. Here we have to calculate the energy of each photon which is given by the Planck’s formula for energy :
\[E = h\nu \]
\[\Rightarrow E = 6.64 \times {10^{ - 34}} \times 1.0 \times {10^{14}}\]
\[\Rightarrow E = 6.64 \times {10^{ - 34}} \times 1.0 \times {10^{14}}\]
\[\therefore E = 6.62 \times {10^{ - 20}}J\]
Thus, the corresponding energy of each photon is \[E = 6.62 \times {10^{ - 20}}J\].

Hence option B is the correct answer.

Note: To determine the corresponding wavelength using the given frequency, we can use \[\nu = \dfrac{c}{\lambda }\]. Wavelength and frequency are inversely proportional to each other. From here, \[\lambda = \dfrac{c}{\nu }\]. If the wavelength of a radiation is given and energy is to be calculated, \[E = \dfrac{{hc}}{\lambda }\] . If all parameters of the equation are in SI units, then the energy from this formula is found by default in joules. This can be converted to eV using the relation, \[1eV = 1.6 \times {10^{ - 19}}J\].