Answer: The wavelength of the flame is 462 nm and color of cesium flame is blue.
Explanation:
To calculate the wavelength, we use Planck's equation, which is:
[tex]E=\frac{hc}{\lambda}[/tex]
where,
E = Energy of 1 photon = [tex]4.30\times 10^{-19}J[/tex]
h = Planck's constant = [tex]6.625\times 10^{-34}J.s[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = ?
Putting values in above equation, we get:
[tex]4.30\times 10^{-19}=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{\lambda }\\\\\lambda=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{4.30\times 10^{-19}}=4.62\times 10^{-7}m=462nm[/tex]
The range of wavelength of blue light lies in range of 500 nm - 435 nm
The calculated wavelength lies in the above range. So, the color of the cesium flame is 462 nm
Hence, the wavelength of the flame is 462 nm and color of cesium flame is blue.
The wavelength of the flame is 462 nm, and the color of cesium flame is blue.
[tex]\bold{By \;plank's \;constant \;E= \dfrac{hc}{\lambda}}[/tex]
where,
Energy of 1 photon (E) = [tex]4.30 \times 10^-^1^9 J[/tex]
Planck's constant (h) = [tex]6.625\times 10^-^3^4 J s[/tex]
Speed of light (c) = [tex]3\times10^8 m/s[/tex]
Wavelength = ?
[tex]\bold{\lambda = \dfrac{6.625\times 10^-^3^4 J s\times3\times10^8m/s}{4.30 \times 10^-^1^9 J} =4.62\times10^-^7 m}[/tex]
= 462 nm
The range of wavelengths of blue light lies in range of 500 nm – 435 n
Thus, The wavelength of the flame is 462 nm, so the color of cesium flame is blue.
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