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# How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

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Solution

## Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n2 = $\infty$). So, the given range of wavelength (380−780 nm) lies in the Balmer series. The wavelength in the Balmer series can be found by $\frac{1}{\lambda }=R\left(\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right)$ Here, R = Rydberg's constant = 1.097×107 m$-$1 The wavelength for the transition from n = 3 to n = 2 is given by $\frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=656.3\mathrm{nm}$ The wavelength for the transition from n = 4 to n = 2 is given by $\frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=486.1\mathrm{nm}$ The wavelength for the transition from n = 5 to n = 2 is given by $\frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{3}=434.0\mathrm{nm}$ The wavelength for the transition from n = 6 to n = 2 is given by $\frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{6}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{4}=410.2\mathrm{nm}$ The wavelength for the transition from n = 7 to n = 2 is given by $\frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{7}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{5}=397.0\mathrm{nm}$ Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5. Lyman series contains wavelengths ranging from 91 nm (for n2 = 2) to 121 nm (n2 =$\infty$). So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series. The wavelength in the Lyman series can be found by $\frac{1}{\lambda }=R\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)$ The wavelength for the transition from n = 2 to n = 1 is given by $\frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=122\mathrm{nm}$ The wavelength for the transition from n = 3 to n = 1 is given by $\frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=103\mathrm{nm}$ The wavelength for the transition from n = 4 to n = 1 is given by $\frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{4}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{3}=97.3\mathrm{nm}$ The wavelength for the transition from n = 5 to n = 1 is given by $\frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{5}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{4}=95.0\mathrm{nm}$ The wavelength for the transition from n = 6 to n = 1 is given by $\frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{6}^{2}}\right)\phantom{\rule{0ex}{0ex}}{\lambda }_{5}=93.8\mathrm{nm}$ So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.  Suggest Corrections  0      Similar questions  Related Videos   Moseley's Law
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