Verified Solution • 2025–26 Session moderate

QJEEM19Chb519: JEE Main 2019 Chemistry - Class 11 JEEM19Chb519 Solution: For any given series of spectral lines of atomic hydrogen, let…

For any given series of spectral lines of atomic hydrogen, let Δv_=\Delta \mathop v\limits^\_ = Δv‾max⁡−Δv‾min⁡\Delta {\overline v _{\max }} - \Delta {\overline v _{\min }} be the difference in maximum and minimum frequencies in cm–1. The ratio Lyman Balmer Δv‾LymanΔv‾Balmer{{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}} is :

Multiple Choice Options

  • A 9:04
  • B 4:01
  • C 27:05:00
  • D 5:04

Step-by-Step Verified Solution

We know,

v‾=RZ2(1n12−1n22)\overline v = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

∴\therefore For Lyman series,

Δv‾Lyman{\Delta {{\overline v }_{Lyman}}} = Δv‾max⁡−Δv‾min⁡{\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}

= [11−1∞]−[11−14]\left[ {{1 \over 1} - {1 \over \infty }} \right] - \left[ {{1 \over 1} - {1 \over 4}} \right]

= 14{{1 \over 4}}

∴\therefore For Balmer series,

Δv‾Balmer{\Delta {{\overline v }_{Balmer}}} = Δv‾max⁡−Δv‾min⁡{\Delta {{\overline v }_{\max }} - \Delta {{\overline v }_{\min }}}

= [14−1∞]−[14−19]\left[ {{1 \over 4} - {1 \over \infty }} \right] - \left[ {{1 \over 4} - {1 \over 9}} \right]

= 19{{1 \over 9}}

∴\therefore Δv‾LymanΔv‾Balmer{{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}} = 1419{{{1 \over 4}} \over {{1 \over 9}}} = 94{{9 \over 4}}

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