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Chemistry Question on Structure of atom

The frequency of radiation emitted when the electron falls from n=4n = 4 to n=1n = 1 in a hydrogen atom will be (Given ionisation energy of H=2.18×1018H = 2.18\times 10^{18} J atom−1atom^{-1} and h=6.625×10−34Js)h = 6.625 \times 10^{-34} Js)

A

1.54×1015s−11.54 \times 10^{15} s^{-1}

B

1.03×1015s−11.03\times 10^{15} s^{-1}

C

3.08×1015s−13.08\times 10^{15} s^{-1}

D

2.00×1015s−12.00\times 10^{15} s^{-1}

Answer

3.08×1015s−13.08\times 10^{15} s^{-1}

Explanation

Solution

The correct answer is C:3.08×1015s−13.08\times 10^{15} s^{-1}
Ionisation energy of H=2.18×10−18Jatom−1H = 2.18 \times 10^{-18} J atom^{-1}
∴          E1\therefore\, \, \, \, \, \, \, \, \, \, E_1 ( Energy of 1st orbit of H-atom)
           =−2.18×10−18Jatom−1\, \, \, \, \, \, \, \, \, \, \, = - 2.18 \times 10^{-18} J atom^{-1}
∴       En=−2.18×10−18n2J atom−1\therefore\, \, \, \, \, \, \, E_n=\frac{-2.18\times10^{-18}}{n^2}J\, atom^{-1}
Z=1Z = 1 for H−atomH-atom
ΔE=E4−E1\Delta E = E_4 - E_1
                  =−2.18×10−1842−−2.18×10−1812\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{-2.18\times10^{-18}}{4^2}-\frac{-2.18\times10^{-18}}{1^2}
                  =−2.18×10−18×[142−112]\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =-2.18\times 10^{-18}\times\Bigg[\frac{1}{4^2}-\frac{1}{1^2}\Bigg]
              ΔE=−2.18×1018×−1516\, \, \, \, \, \, \, \, \, \, \, \, \, \, \Delta E=-2.18\times 10^{18}\times -\frac{15}{16}
                  =+2.0437×10−18J atom−1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =+2.0437\times 10^{-18} J\, atom^{-1}
∴          v=ΔEh\therefore\, \, \, \, \, \, \, \, \, \, v=\frac{\Delta E}{h}
                  =2.0437×10−18J atom−16.6256×10−34J s\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{2.0437 \times 10^{-18} J\, atom^{-1}}{6.6256\times 10^{-34}J\, s}
                  =3.084×1015s−1 atom−1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =3.084 \times 10^{15} s^{-1}\, atom^{-1}