A photon of wavelength 656.3nm is emitted in the Balmer series of a Hydrogen emission lamp. (a). Show that the frequency of the photon is 4.57*10^14 Hz. (b).Use the Planck-Einstein relationship to calculate the energy of the photon.

(a). Firstly, pull out useful information . λ=656.3nm Secondly, identify relevant equation and rearrange to find frequency. λ=c/f => f=c/λ Calculate frequency using the given values (c=310^8 m/s), ensuring to convert nm into m f=c/λ=(310^8)/(656.310^-9) f=4.5710^14 Hz => as required, ensuring correct units (Hz)(b). Firstly, identify relevant equation (Planck-Einstein Relationship) and required variables. f=4.5710^14 Hz h=6.6310^-34 Js => Planck's Constant E=hf Then, calculate energy using the above equation E=hf=(6.6310^-34)(4.5710^14) E= 3.0310^-19J => Final answer, with correct units Joules



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Answered by Robert M. Physics tutor

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