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balu736 [363]
3 years ago
15

When sodium is excited in a flame, two ultraviolet spectral lines at lambda - 372.1 nm and lambda = 376.4 nm respectively are em

itted . which wavelength represented photons?
 a) higher energy?
b) longer wavelengths?
c) higher frequences?
Chemistry
2 answers:
Allushta [10]3 years ago
6 0

Answer: Wavelength which represented photons with :

a) Higher energy is 372.1 nm

b) Longer wavelength is 376.4 nm

c) Higher frequency is 372.1 nm

Explanation:

E=h\nu =\frac{hc}{\lamda } (Planck's equation)

h=\text{Planck's constant}=6.62\times 10^{-34}J-s,c=\text{speed of light}=3\times 10^{8}m/s

\lambda = wavelength of the photon with energy E in meters.

\nu = frequency of the photon with energy E in hertz.

For first spectral line:

\lambda _1=372.1 nm=372.1\times 10^{-9} m(1nm=1\times 10^{-9} m)

E_1=\frac{hc}{\lambda_1}=\frac{(6.62\times 10^{-34}J-s)(3\times 10^{8}m/s)}{372.1\times 10^{-9} m}=5.33\times 10^{-19} joules

\nu _1=\frac{c}{\lambda _1}=\frac{3\times 10^{8}m/s}{372.1\times 10^{-9} m}=0.08 Hertz

For second spectral line:

\lambda _2=376.4 nm=376.4\times 10^{-9} m(1nm=1\times 10^{-9} m)

E_2=\frac{hc}{\lambda_2}=\frac{(6.62\times 10^{-34}J-s)(3\times 10^{8}m/s)}{376.4\times 10^{-9} m}=5.27\times 10^{-19} joules

\nu _2=\frac{c}{\lambda _2}=\frac{3\times 10^{8}m/s}{376.4\times 10^{-9} m}=0.07 Hertz

suter [353]3 years ago
3 0
The wavelength that represented photons is at <span>lambda = 376.4 nm.
The wavelength with the higher energy is at </span><span>lambda - 372.1 nm.
The longer wavelength is, of course, </span> at lambda 376.4 <span>nm.
The wavelength with the higher frequency is </span> 376.4 <span>nm.</span><span>
</span><span>
</span>
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