Question 490: Atoms & Nuclei

Question

The wavelength of light for the least energetic photons emitted in the Lyman series of the hydrogen spectrum is nearly [Take $$\mathrm{hc}=1240 ~\mathrm{eV}$$ - $$\mathrm{nm}$$, change in energy of the levels $$=10.2 ~\mathrm{eV}$$ ]

Accepted Answer

For least energetic photon emitted in Lyman series, $$E=E_2-E_1=10.2 \mathrm{~eV}$$
$$\begin{aligned}
\lambda=\frac{\mathrm{hc}}{\mathrm{E}} & =\frac{1240}{10.2} \mathrm{~nm} \
& =121.57 \mathrm{~nm} \approx 122 \mathrm{~nm}
\end{aligned}$$

Answer

$$150 \mathrm{~nm}$$

Answer

$$122 \mathrm{~nm}$$

Answer

$$102 \mathrm{~nm}$$

Answer

$$82 \mathrm{~nm}$$

Atoms & Nuclei

The wavelength of light for the least energetic photons emitted in the Lyman series of the hydrogen spectrum is nearly [Take $\mathrm{hc}=1240 ~\mathrm{eV} - \mathrm{nm}, change in energy of the levels =10.2 ~\mathrm{eV}$ ]

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