Asked by Anonymous
High-pressure sodium vapor lamps are used in street lighting. The two brightest lines in the sodium spectrum are at 589.00 and 589.59 nm. What is the diffrence in energy per photon of the radiations corresponding to these two lines?
Answers
Answered by
bobpursley
E=planksconstant*(c/lambda)
= k*c (1/589nm -1/589.59nm)
= k*c*(589.59-589)/589*589.59
= k*c*(.59E9)
= 6.6E-34*3E8(.59E9)
about 1.1E-16 Joule check that math.
= k*c (1/589nm -1/589.59nm)
= k*c*(589.59-589)/589*589.59
= k*c*(.59E9)
= 6.6E-34*3E8(.59E9)
about 1.1E-16 Joule check that math.
Answered by
Anonymous
the correct answer is 3.375x10^-22 J. But i don't know how to get this answer correctly.
Answered by
bobpursley
E=planksconstant*(c/lambda)
= k*c (1/589nm -1/589.59nm)
= k*c*(589.59-589)/589*589.59 1/nm
= k*c*( 1.69897351e-6)1E9
= 6.6E-34*3E8( 1.69897351e-6)*1e9
3.3639675e-22
= k*c (1/589nm -1/589.59nm)
= k*c*(589.59-589)/589*589.59 1/nm
= k*c*( 1.69897351e-6)1E9
= 6.6E-34*3E8( 1.69897351e-6)*1e9
3.3639675e-22
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