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  • Energy E of a hydrogen atom with principal quantum number n is given by <img alt="\mathrm{E=13.6/n^{2}eV}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE%3D13.6/n%5E%7B2%7DeV%7D

Energy E of a hydrogen atom with principal quantum number n is given by \mathrm{E=13.6/n^{2}eV}. The energy of a photon ejected when the electron jumps from \mathrm{n=3} state to \mathrm{n=2} state of hydrogen is approximately
 

Option: 1

\mathrm{ 1.5 eV}
 


Option: 2

\mathrm{ 0.85 eV}
 


Option: 3

\mathrm{ 3.3 \: eV}
 


Option: 4

\mathrm{1.9\: eV}


Answers (1)

best_answer

Energy of ejected photon,

\mathrm{E =R h c\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) }

\mathrm{ =13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \mathrm{eV} }

\mathrm{ =13.6 \times \frac{5}{36}=1.9 \mathrm{eV}}

Hence option 4 is correct.



 

Posted by

Ritika Jonwal

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