Question bank/NEET UG/Chemistry

NEET UGChemistryModerateNEET 2025

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

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Sol. ΔE=hcλ=Efinal−Einitial(En=−RHn2)\Delta E = \frac{hc}{\lambda} = E_{final} - E_{initial} \left(E_n = \frac{-R_H}{n^2}\right)

ΔE2→3=hcλ2→3=E3−E2=−RH32−(−RH22)\Delta E_{2\rightarrow 3} = \frac{hc}{\lambda_{2\rightarrow 3}} = E_3 - E_2 = \frac{-R_H}{3^2} - \left(\frac{-R_H}{2^2}\right)

=RH(14−19)= R_H\left(\frac{1}{4} - \frac{1}{9}\right)

=RH×536= R_H \times \frac{5}{36}

∴λ2→3=hc⋅36RH⋅5\therefore \lambda_{2\rightarrow 3} = \frac{hc \cdot 36}{R_H \cdot 5}

ΔE4→6=E6−E4=−RH36−−RH16=RH×2036×16\Delta E_{4\rightarrow 6} = E_6 - E_4 = \frac{-R_H}{36} - \frac{-R_H}{16} = \frac{R_H \times 20}{36 \times 16}

hcλ4→6=RH×2036×16\frac{hc}{\lambda_{4\rightarrow 6}} = \frac{R_H \times 20}{36 \times 16}

λ4→6=hc×36×16RH⋅20\lambda_{4\rightarrow 6} = \frac{hc \times 36 \times 16}{R_H \cdot 20}

λ2→3λ4→6=hc⋅36RH⋅5hc×36×16RH⋅20\frac{\lambda_{2\rightarrow 3}}{\lambda_{4\rightarrow 6}} = \frac{\frac{hc \cdot 36}{R_H \cdot 5}}{\frac{hc \times 36 \times 16}{R_H \cdot 20}}

=14= \frac{1}{4}

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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