Electromagnetic radiation and spectra

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Question

A yellow coloured spectral line was observed at $5890 A^{o}$ in the spectrum of a sodium vapour lamp. Calculate the minimum accelerating potential that is required to excite this line in an electron tube containing sodium vapour

Moderate

Solution

$λ = 5890 \times 10^{- 10} m; c = 3.0 \times 10^{8} {ms}^{-}$
We know that $E = h v; E = \frac{h c}{λ}$
$E = \frac{6.626 \times 10^{- 34} J . s \times 3.0 \times 10^{8} {ms}^{-}}{5890 \times 10^{- 10} m}$
$= 3.37 \times 10^{- 19} J$
But, $1 eV = 1.6 \times 10^{- 19} J$
Hence, $E = 3.37 \times 10 - 19 J \times \frac{1 eV}{1.6 \times 10^{- 19} J}$
$= 2.1 eV$
= 2.1 V x charge on one electron
$∴$ 2.1 Vis needed to accelerate the electron away from the atom.

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Electromagnetic radiation and spectra

Unlock the full solution & master the concept.

A yellow coloured spectral line was observed at 5890Ao in the spectrum of a sodium vapour lamp. Calculate the minimum accelerating potential that is required to excite this line in an electron tube containing sodium vapour

λ=5890×10−10m;c=3.0×108ms−We know that E=hv;E=hcλE=6.626×10−34J.s×3.0×108ms−5890×10−10m=3.37×10−19JBut, 1eV=1.6×10−19JHence, E=3.37×10−19J×1eV1.6×10−19J=2.1eV= 2.1 V x charge on one electron∴ 2.1 Vis needed to accelerate the electron away from the atom.

Ready to Test Your Skills?

free study material

A yellow coloured spectral line was observed at $5890 A^{o}$ in the spectrum of a sodium vapour lamp. Calculate the minimum accelerating potential that is required to excite this line in an electron tube containing sodium vapour