Question

If the ionisation potential of the hydrogen atom is $13.6\mathrm{eV}$, then what will be the longest wavelength needed to remove an electron from the $1^{\text{st }}$ Bohr's orbit of ${\mathrm{He}}^{+}$ ion?

• (A) $2.284\times {10}^{-10}\mathrm{m}$
• (B) $2.284\times {10}^{-8}m$
• (C) $228.4\mathrm{A}$
• (D) Both (B) and (C)

Answer 1

The Correct Option is D

Explanation:
Given:Ionisation potential of the H-atom $=13.6eV$We have to find out the longest wavelength needed to remove an electron from the first Bohr's orbit of He+ ion.According to Bohr's model of an atom, the energy of a hydrogen-like ion is given as:$E_n=-2.18\times {10}^{-18}\left(\frac{Z^2}{n^2}\right)J/$ atomwhere Z = Nuclear charge (equal to atomic number )n = No. of the orbitIonisation potential (I.P) is given as:$I\cdot P=E_{\mathrm{\infty }}-E_n$$=0-\left[\begin{array}{lll}-2.18& \times & {10}^{-18}\end{array}\left(\frac{Z^2}{n^2}\right)\right]J$$=2.18\times {10}^{-18}\left(\frac{Z^2}{n^2}\right)J/$ atom.....(i)For hydrogen atom, I.P. $=13.6\mathrm{eV}$$=13.6\times 1.6\times {10}^{-19}J$$(As,1eV=1.6\times {10}^{-19}J)$$=21.76\times {10}^{-19}J$For He+ ion,Number of the orbit, n = 1Atomic number, Z of He+ ion = 2Substituting values in equation (i), we getI.E $=2.18\times {10}^{-18}\left(\frac{2^2}{1^2}\right)J$$=8.72\times {10}^{-18}J$Now,According to Planck's quantum theory of radiation, energy is givenas:$E=\frac{hc}{\lambda }$....(ii)where, $h=$ Planck's constant $=6.63\times {10}^{-34}Js$$\mathrm{c}=$ Speed of light $=3\times {10}^8\mathrm{m}/\mathrm{s}$$\lambda =$ WavelengthSubstituting the values in equation (ii), we get$8.72\times {10}^{-18}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{\lambda }$${\lambda }_{max}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{8.72\times {10}^{-18}}$$=2.2809\times {10}^{-8}\mathrm{m}$$=228.09\times {10}^{-10}\mathrm{m}$ÅÅ$=228.09\AA (As,1\AA ={10}^{-10}m)$Hence, the correct option is (D).