calculate the energy of one photon of light at wavelength equal to the major line emission of terbium at 547 nm.



Answer :

The energy of one photon is E = 3.634 J.

Wavelength = [tex]547 * 10^{-9}[/tex] m

Energy , E = hc / λ

             E = 6.626 [tex]* 10^{-34} Js[/tex] * 3 [tex]* 10^{8} m /s[/tex] / 547 * [tex]10^{-9} m[/tex]

             E = 3.634 J

Energy is the quantitative attribute that is conveyed to a body or a physical system and is perceptible as heat and light as well as in the performance of work. Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or destroyed. The joule is the SI's (International System of Units) unit of measurement for energy (J).

The distance between consecutive corresponding points of the same phase on the wave, such as two adjacent crests, troughs, or zero crossings, is known as the wavelength of a periodic wave; it is the length over which the wave's shape repeats.  It is a property of both traveling waves and standing waves, as well as other spatial wave patterns.  The inverse of the wavelength is known as the spatial frequency. The Greek letter lambda (λ) is frequently used to represent wavelength.

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