Physical Chemistry> pl urgent help

pl urgent help

Akhil Verma , 9 Years ago

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6 Answers

Ankit Sharma

Last Activity: 9 Years ago

THE ANSWER IS C)3

as both the hydrogen atoms are identical

so they will give same balmer lines so we will count balmer lines of only one atom

i.e

5-2

4-2

3-2

these two wavelength can only be balmer one and visible one according to me

AND ALSO 4TH EXCITED STATE = 5TH ORBIT AS

2ND ORBIT = 1ST EXCITED STATE

AS WE DONT COUNT GROUND STATE IN EXCITED STATE LIST !

hope that it is useful !

approve if useful !

Akhil Verma

Last Activity: 9 Years ago

@ankit sharma the anser is given 2 :(

Ankit Sharma

Last Activity: 9 Years ago

but hav u found any flaw in my answer

the answer in the book or site u refered might be wrong as the might have taken fourth excited state as fourth orbit or any other as i hav given my logical explaination and if u find any flaw then plz tell me otherwise dont dissapprove as the question setter can also be wrong akhil

so hav u found any flay in my reasoning then tell !!

otherwise approve if it helps !!

the main motto is to understand the concept and give correct reasoning thats all !!!

who cares that the answer given is write or wrong !!

u should hav confidence on urself and ur answer !

i hav and i know i m write thats all !!!

noogler

Last Activity: 9 Years ago

ankit u told there are 3ways i.e, 5-2,4-2,3-2

but what about 5-4-3-2,5-4-2,5-3-2

y u didnt considered them plz explain

Akhil Verma

Last Activity: 9 Years ago

@ankit

answer the question this guy @manam if your logic is correct y not des passibilities and the answer is not wrong read the question again it is about 2 hydrogen so we cant a consider the genral method of jumping of elctrons from 1 orbital to other orbital and caluclating the lines