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2.19    The electron energy in hydrogen atom is given by E_n = (-2.18\times 10^{-18})/n^2 \ \textup{J}. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

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The expression for the energy of an electron in hydrogen is:

E_{n}=-\frac{2n^2me^4Z^2}{n^2h^2}

Where m is mass of electrons, Z is the atomic mass of an atom, e is the charge of an electron, and h is the Planck's constant.

and electron energy in the hydrogen atom is given by,

E_{n}=-\frac{(2.18\times10^{-18})}{n^2}J

The electron energy in n=2 orbit is:

E_{n}=-\frac{(2.18\times10^{-18})}{2^2}J = 0.5465\times10^{-18}J

Therefore, the energy required for the ionization from n=2 is 5.45\times10^{-19}J

Now, the longest wavelength of light that can be used to cause this transition will be:

E=\frac{hc}{\lambda} 

\lambda = \frac{hc}{E} =\frac{(6.626\times10^{-34}Js)(3\times10^8)m/s}{5.45\times10^{-19}J}

    =3.674\times10^{-7}m = 3.674\times10^{-5}cm

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Divya Prakash Singh

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