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`        The series limit for the Paschan series of hydrogen spectrum occurs at 8205.8A.CalculateA) ionization energy of the hydrogen atomB)wave length of the photon that would remove the electron the electron in the ground state of m the hydrogen atom `
one month ago

## Answers : (2)

Arun
23032 Points
```							For a hydrogen molecule, made out of a circling electron bound to a core of one proton, an ionization vitality of2.18 × 10 −18 joule (13.6 electron volts) is required to drive the electron from its most reduced vitality level completely out of the iota. λ=8205.8A The Rydberg Expression gives: 1λ=R[1n21−1n22] R is the Rydberg Constant = 1.097×10−2nm−1 n1=2 n2=4 These allude to the guideline quantum quantities of the vitality levels which the electrons move from. 1λ=1.097×10−2[0.25−0.0625]=0.00206nm−1 λ=485.4nm The sub - shells d and p given are unimportant to the question as, in the hydrogen particle, the sub - shells inside a vitality level are deteriorate
```
one month ago
Vikas TU
9499 Points
```							Dear student The above explanaton is not correct , Please follow this i).The energy corresponding to 8205.8 Angestrum will be del E = hc/Lambda put all the corresponding values del E = 2.422 * 10^(-19) J Convert this into eV delE = 1240 /820.58 = 1.512 eV Now  Ionization energy of hydrogen atom E will be calculated del E = E * Z^2 (1/n1^2 - 1/n2^^2) n1 =3 n2 = infinity Z =1 put the value of del E the corresponding value of E will be E = 13.608 eV ii) For wavelength Put Lambda = hc/E lambda = 91.6 nm Hope this helps
```
one month ago
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