First slide

Classical atomic models

Question

A hydrogen atom in its ground state is irradiated by light of wavelength 970 $\overset{o}{A}$ . Taking hc/e = 1.237 × 10^-6 eV m and the ground state energy of hydrogen atom as -13.6 eV, the number of lines present in the emission spectrum is

Moderate

Solution

Absorbed quantum energy in (eV) $= \frac{hc}{eλ}$
$= \frac{1.237 \times 10^{- 6}}{970 \times 10^{- 10}}$
= 12.75 eV
Energy of nth shell =-13.6 + 12.75
= -0.85 eV
We know,
$E_{n} = \frac{E_{1}}{n^{2}}$
$∴ n^{2} = \frac{E_{1}}{E_{n}} = \frac{- 13 .6}{- 0 .85} = 16$
i.e., n = 4.
Number of spectral emission lines $= \frac{n (n - 1)}{2} = \frac{4 \times 3}{2} = 6$

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