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A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? |
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Answer» Solution : Ground state energy of hydrogen gas at room TEMPERATURE = - 13.6 eV. When the energy BEAM used to bombard gaseous hydrogen then its energy becomes = - 13.6 + 12.5 eV = -1.1 eV. Now energy in nth orbit `E_(n)=-(13.6)/(n^(2))eV` `:.-1.1eV=(13.6)/(n^(2))eV` `:. n^(2)=(13.6)/(1.1)=12.36` `:. n=3.51` but possibel orbits n=3  `:.` Spectral line of emiision spectra `=(n(n-1))/(2)` `=(3(-1))/(2)` =3 Hence in transition from n = 3 to n = 1 an n = 2 to n = 1 two lines of Lyman series and i transition n = 3 to n = 2, one line of Balme series are obtained. `RARR` In transition from n = 3 to n = 1 th wavelength of `beta`-line emitted in Lyman series. `lambda_(31)=(HC)/(E_(1)-E_(3))=(6.62xx10^(-34)xx3xx10^(8))/((-13.6-(-1.51))` `:. lambda_(31)=(19.875xx10^(-26))/(12.09xx1.6xx10^(-19))` `=1.0274xx10^(-7)m` `~~103xx10^(-9)m=103 m` In transition from n = 2 to n = 1, the wavelength of c-line emitted in Lyman series, `lambda_(21)=(hc)/(E_(1)-E_(2))=(6.625xx10^(-34)xx3xx10^(8))/([-13.6-(3.4)]eV)` `=(19.875xx10^(-26))/(10.2xx1.6xx10^(-19))` `=1.2178xx10^(-7)m` `~~122xx10^(-9)m=122 mm` `rArr` In transition from n = 3 to n = 2. wavelength of a-line emitted in Balmer series, `lambda_(32)=(hc)/(E_(2)-E_(3))=(6.625xx10^(-34)xx3xx10^(8))/([-3.4-(-1.51)]eV)` `:. lambda_(32)=(19.875xx10^(-26))/(1.89xx1.6xx10^(-19))` `6.5724xx10^(-7)m` `=657xx10^(-9)m` `657nm` |
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