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PLEASE HELP!!!!! An electron jumps from one energy level to another in an atom radiating 4.5*10^-19 Joules. If Planck's constant is 6.63*10^-34Js, what is the wavelength of the radiation? (C =3*10^8) 2)

User Milind
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Answer: λ = 4.42 × 10^-7 metres

Step-by-step explanation:

The emitting radiation energy is the change in energy dE. Where

dE = 4.5×10^-19J

Planck's constant h = 6.63*10^-34Js,

Speed of light C = 3×10^8

The wavelength(λ) of the radiation can be calculated by using the formula

(λ) = hC/dE

Substitute all the parameters into the formula above

(λ) = (6.63×10^-34 × 3×10^8)/4.5×10^-19

(λ) = 1.989^-25/4.5×10^-19

(λ) = 4.42×10^-7 m

Therefore, the wavelength (λ) of the radiation is 4.42 × 10^-7 metres

User Slwr
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