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Chemistry Question on Structure of atom

Let the energy of an nthn^{th} orbit of HH atom be −21.76×1019/n2 J-21.76 \times 10^{19}/n^2\, J . What will be the longest wavelength of energy required to remove an electron from the third orbit?

A

0.628 nm0.628\, nm

B

1.326×10−7 m1.326 \times 10^{-7}\, m

C

0.798 pm0.798\, pm

D

0.821 μm0.821 \,\mu m

Answer

0.821 μm0.821 \,\mu m

Explanation

Solution

En=−21.76×10−19n2JE_{n}=\frac{-21.76\times10^{-19}}{n^{2}} J
For n=3n = 3
En=−21.76×10−19(3)2E_{n}=\frac{-21.76\times10^{-19}}{\left(3\right)^{2}}
=−2.42×10−19J=-2.42\times10^{-19} J
E=hcλE=\frac{hc}{\lambda}
⇒λ=hcE\Rightarrow \lambda=\frac{hc}{E}
=6.6×10−34×3×1082.42×10−19=\frac{6.6\times10^{-34}\times3\times10^{8}}{2.42\times10^{-19}}
=8.2×10−7 m=8.2\times10^{-7}\,m
=0.82×10−6 m=0.82\times10^{-6}\,m
=0.82 μm=0.82\, \mu m