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Grade: 12
        The two longest wavelengths of Balmer series of triply ionized beryllium (z=4) are:A)41nm B)30.4nm C)45nm D)39nmIt is a multi correct answer question. Please answer this it is very URGENT
2 years ago

Answers : (2)

Arun
23047 Points
							
Dear student
 
Since the radiation of the Balmer series ends with n= 2.
Hence two longest wavelengths are the radiations corresponding to 
n = 3 to n= 2
And n = 4 to n =2
 
Now  find (E3- E2) = - (13.6 ev) 4² ( 1/9 -  1/4)
Then
Lambda = h c /E
Similarily for
(E4- E2)
 
Hope you understand.
 
Regards
Arun (askIITians forum expert)
2 years ago
kipkoech patrick
15 Points
							
1 first wavelength ​
n = 3  ,  E = _ 13.6 eV4 (1/9 – 1/4) 
but lamda = hc/E 
This gives the first longest wavelength . 
for  the second longest wavelenthg 
take n = 4 
therefore E = -13.6 eV 42 (1/16 – ¼)
Then lamda = hc/E
This gives the second wavelenthg
 
8 months ago
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