Hydrogen gas absorbs light of wavelength 103 nm. Afterward, what wavelengths are seen in the emission spectrum?

arbixerwoxottdrp1l
2022-04-07
Answered

Hydrogen gas absorbs light of wavelength 103 nm. Afterward, what wavelengths are seen in the emission spectrum?

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Lea Johnson

Answered 2022-04-08
Author has **13** answers

Step 1

Let the absorbed wavelength by the hydrogen gas can be

$\lambda =103\text{}nm$

Therefore, the energy absorbed is

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{(6.625\times {10}^{-34})(3\times {10}^{8})}{(103\times {10}^{-9})}\times (\frac{1}{1.6\times {10}^{-19}})eV\phantom{\rule{0ex}{0ex}}=12.05\text{}eV\phantom{\rule{0ex}{0ex}}\therefore E=12.05\text{}eV$

And the ground state energy of the hydrogen is -13.6 eV

Step 2

Therefore, to find the value of n we will use the following method

$(\frac{-13.6}{{n}^{2}})-(\frac{-13.6}{{1}^{2}})=12.05\phantom{\rule{0ex}{0ex}}\Rightarrow n=\sqrt{\frac{13.6}{1.55}}\phantom{\rule{0ex}{0ex}}=2.96\phantom{\rule{0ex}{0ex}}\approx 3\phantom{\rule{0ex}{0ex}}\therefore n=3$

Step 3

Therefore, the electron is jumping from n =3 to n = 1 and the corresponding wave lengths can be calculated as follows

Now the line corresponding to ${n}_{2}=3$ to ${n}_{1}=1$ is the first line , its wave length is maximum

That is,

$\frac{1}{{\lambda}_{max}}=R(\frac{1}{{1}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}=(1.1\times {10}^{7})(1-\frac{1}{9})\phantom{\rule{0ex}{0ex}}=0.977\times {10}^{7}\phantom{\rule{0ex}{0ex}}\therefore {\lambda}_{max}=1022\text{}\stackrel{\u02da}{A}$

Step 4

Similarly, the minimum wavelength is

$\frac{1}{{\lambda}_{min}}=R(\frac{1}{{1}^{2}}-\frac{1}{{\mathrm{\infty}}^{2}})\phantom{\rule{0ex}{0ex}}=(1.1\times {10}^{7})(1)\phantom{\rule{0ex}{0ex}}=1.1\times {10}^{7}\phantom{\rule{0ex}{0ex}}=912\phantom{\rule{0ex}{0ex}}\therefore {\lambda}_{min}=912\text{}\stackrel{\u02da}{A}$

Let the absorbed wavelength by the hydrogen gas can be

$\lambda =103\text{}nm$

Therefore, the energy absorbed is

$E=\frac{hc}{\lambda}\phantom{\rule{0ex}{0ex}}=\frac{(6.625\times {10}^{-34})(3\times {10}^{8})}{(103\times {10}^{-9})}\times (\frac{1}{1.6\times {10}^{-19}})eV\phantom{\rule{0ex}{0ex}}=12.05\text{}eV\phantom{\rule{0ex}{0ex}}\therefore E=12.05\text{}eV$

And the ground state energy of the hydrogen is -13.6 eV

Step 2

Therefore, to find the value of n we will use the following method

$(\frac{-13.6}{{n}^{2}})-(\frac{-13.6}{{1}^{2}})=12.05\phantom{\rule{0ex}{0ex}}\Rightarrow n=\sqrt{\frac{13.6}{1.55}}\phantom{\rule{0ex}{0ex}}=2.96\phantom{\rule{0ex}{0ex}}\approx 3\phantom{\rule{0ex}{0ex}}\therefore n=3$

Step 3

Therefore, the electron is jumping from n =3 to n = 1 and the corresponding wave lengths can be calculated as follows

Now the line corresponding to ${n}_{2}=3$ to ${n}_{1}=1$ is the first line , its wave length is maximum

That is,

$\frac{1}{{\lambda}_{max}}=R(\frac{1}{{1}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}=(1.1\times {10}^{7})(1-\frac{1}{9})\phantom{\rule{0ex}{0ex}}=0.977\times {10}^{7}\phantom{\rule{0ex}{0ex}}\therefore {\lambda}_{max}=1022\text{}\stackrel{\u02da}{A}$

Step 4

Similarly, the minimum wavelength is

$\frac{1}{{\lambda}_{min}}=R(\frac{1}{{1}^{2}}-\frac{1}{{\mathrm{\infty}}^{2}})\phantom{\rule{0ex}{0ex}}=(1.1\times {10}^{7})(1)\phantom{\rule{0ex}{0ex}}=1.1\times {10}^{7}\phantom{\rule{0ex}{0ex}}=912\phantom{\rule{0ex}{0ex}}\therefore {\lambda}_{min}=912\text{}\stackrel{\u02da}{A}$

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