An electron in an excited hydrogen atom starts in n=4 and emits a 486 nm photon

Question

An electron in an excited hydrogen atom starts in n=4 and emits a 486 nm photon as it relaxes to a final energy level, n. a. What is the frequency of the photon emitted during this transition? b. What is the energy of the photon emitted during this transition? c. What is the difference in the two energy levels of this transition? Watch your signs! d. Determine the end (final) value of n in this transition. An electron in an excited hydrogen atom starts in n=4 and emits a 486 nm photon as it relaxes to a final energy level, n. An electron in an excited hydrogen atom starts in n=4 and emits a 486 nm photon as it relaxes to a final energy level, n. a. What is the frequency of the photon emitted during this transition? b. What is the energy of the photon emitted during this transition? c. What is the difference in the two energy levels of this transition? Watch your signs! d. Determine the end (final) value of n in this transition.

Explanation / Answer

wavelength =c/f..........

f = c/lamda=6.172*10^14 Hz.............

b)E =h*f = 40.88*10-^-17 J......

c) E-En = 40.88*10-^-17 J......

d)

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