The shortest wavelength of H atom in the Lyman series is λ₁. The longest wavelength in the Balmer Series of H⁺ is:
The shortest wavelength of H atom in the Lyman series is λ₁. The longest wavelength in the Balmer Series of H⁺ is:
For hydrogen atom :
For Lyman series n? = 1 & n? = ∞
1/λH = RH [1/1² - 1/∞²] So, λ = 1/RH
For He? ion Balmer series n? = 2 & n? = 3
1/λHe? = RH * Z² [1/n? ² - 1/n? ²]
1/λHe? = RH * 4 * [1/4 - 1/9] = RH * 4 * (5/36)
1/λHe? = (5/9)RH = (5/9) (1/λ)
(λHe? ) = (9/5)λ
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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