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?
the correct answer is 3.375x10^-22 J. But i don't know how to get this answer correctly.
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.
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
To find the difference in energy per photon between the two lines in the sodium spectrum, we can use the equation:
ΔE = hc/λ
Where:
ΔE is the difference in energy per photon,
h is Planck’s constant (6.62607015 × 10^-34 J·s),
c is the speed of light in a vacuum (2.998 × 10^8 m/s),
and λ is the wavelength of light.
First, we need to convert the wavelengths of the two lines into meters. We can do this by dividing the given values by 10^9 (since 1 nm = 10^9 m):
λ1 = 589.00 nm = 589.00 × 10^-9 m
λ2 = 589.59 nm = 589.59 × 10^-9 m
Now, we can calculate the difference in energy per photon:
ΔE = (hc/λ2) - (hc/λ1)
Substituting the known values:
ΔE = [(6.62607015 × 10^-34 J·s × 2.998 × 10^8 m/s) / (589.59 × 10^-9 m)] - [(6.62607015 × 10^-34 J·s × 2.998 × 10^8 m/s) / (589.00 × 10^-9 m)]
Simplifying the equation yields:
ΔE = (3.558 × 10^-19 J) - (3.553 × 10^-19 J)
Therefore, the difference in energy per photon of the radiations corresponding to the two lines is approximately 5 × 10^-22 J (joules).