Whenever a photon is emitted by hydrogen in balmer series it is followed by another in lyman series what wavelength does this latter photon correspond

Correct Answer - `5.8 cm`

Correct Answer - `4.26 xx 10^(-16) J`

Correct Answer - `1379 Å`

Correct Answer - (i) 1.1 eV (ii) `1.6 xx 10^(-6) A` ; unchanged

Correct Answer - `0.85 eV, 3.4 eV, 6.16 xx 10^(14) Hz, 4.87 xx 10^(-7)m`; yes

Correct Answer - 1215.14 Å

Correct Answer - 5893 Å

Correct Answer - `[11.2eV]`

Correct Answer - 3.25 m/s

Given :  ni = 5, nf = 2 To find :  Wavelength of the photon emitted Formulae : i. \(\bar v\) = 109677 [\(\frac{1}{n_f^2}\) - \(\frac{1}{n_i^2}\)] cm-1 ii. λ = \(\frac{1}{\bar v}\)  Calculation :  According to Rydberg equation [formula (i)], \(\bar v\) = 109677 [\(\frac{1}{n_f^2}\) -...