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Write Brief Answer · Q36

Q.The quantum mechanical treatment of the hydrogen atom gives the energy value: En=−13.6n2 eV atom−1E_n=\dfrac{-13.6}{n^2}\ \text{eV atom}^{-1}.

(i) Use this expression to find ΔE\Delta E between n=3n=3 and n=4n=4.
(ii) Calculate the wavelength corresponding to the above transition.
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Step 1. E3=−13.632=−13.69=−1.511E_3=\dfrac{-13.6}{3^2}=\dfrac{-13.6}{9}=-1.511 eV; E4=−13.642=−13.616=−0.850E_4=\dfrac{-13.6}{4^2}=\dfrac{-13.6}{16}=-0.850 eV.

Step 2. ΔE=E4−E3=−0.850−(−1.511)=0.661\Delta E=E_4-E_3=-0.850-(-1.511)=0.661 eV (the energy that must be absorbed to go from n=3n=3 to n=4n=4).

Step 3. Convert to joules: ΔE=0.661×1.602×10−19=1.059×10−19\Delta E=0.661\times1.602\times10^{-19}=1.059\times10^{-19} J.

Step 4. Use λ=hc/ΔE\lambda=hc/\Delta E: λ=(6.626×10−34)(3×108)1.059×10−19≈1.877×10−6 m\lambda=\frac{(6.626\times10^{-34})(3\times10^8)}{1.059\times10^{-19}}\approx1.877\times10^{-6}\ \text{m} …

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