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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

Naga sree harsha reddy , 7 Years ago
Grade 12
anser 2 Answers
Arun

Last Activity: 7 Years ago

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)

kipkoech patrick

Last Activity: 6 Years ago

1 first wavelength ​
n = 3  ,  E = _ 13.6 eV4 2 (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
 

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