Question

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

• A. (A) 2.284×10−10 m
• B. (B) 2.284×10−8m
• C. (C) 228.4 A
• D. (D) Both (B) and (C)

Answer 1

Answer: (D)

(D) Both (B) and (C)

Answer 2

Explanation:
Given:Ionisation potential of the H-atom =13.6eVWe 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:En=−2.18×10−18(Z2n2)J/ atomwhere Z = Nuclear charge (equal to atomic number )n = No. of the orbitIonisation potential (I.P) is given as:I⋅P=E∞−En=0−[−2.18×10−18(Z2n2)]J=2.18×10−18(Z2n2)J/ atom.....(i)For hydrogen atom, I.P. =13.6eV=13.6×1.6×10−19J(As,1eV=1.6×10−19J)=21.76×10−19JFor He+ ion,Number of the orbit, n = 1Atomic number, Z of He+ ion = 2Substituting values in equation (i), we getI.E =2.18×10−18(2212)J=8.72×10−18JNow,According to Planck's quantum theory of radiation, energy is givenas:E=hcλ....(ii)where, h= Planck's constant =6.63×10−34Jsc= Speed of light =3×108m/sλ= WavelengthSubstituting the values in equation (ii), we get8.72×10−18=6.63×10−34×3×108λλmax=6.63×10−34×3×1088.72×10−18=2.2809×10−8m=228.09×10−10mÅÅ=228.09Å(As,1Å=10−10m)Hence, the correct option is (D).