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The series limit for the Paschan series of hydrogen spectrum occurs at 8205.8A. Calculate A) ionization energy of the hydrogen atom B)wave length of the photon that would remove the electron the electron in the ground state of m the hydrogen atom

The series limit for the Paschan series of hydrogen spectrum occurs at 8205.8A.
Calculate
A) ionization energy of the hydrogen atom
B)wave length of the photon that would remove the electron the electron in the ground state of m the hydrogen atom
 

Grade:11

2 Answers

Arun
25750 Points
4 years ago
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
Vikas TU
14149 Points
4 years ago
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 

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