d2

JEE Mains 2015 Papers › Physical Chemistry questions › d2

#Chemistry #Physical Chemistry

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6 Answers

6

nix_13 12 ·2009-09-26 22:28:40

16,17,18

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106

Asish Mahapatra ·2009-09-26 22:36:04

17. radius of first Bohr orbit is x (im assuming H-atom)

wavelength of third orbit,

Î»=h/mv₃

=> mv₃ = h/Î»

now, mv₃r₃ = 3h/2Î

so, r₃/Î» = 3/2Î ... (i)

Now r is directly proportional to n²/Z.

=> r₃/x = 3² = 9
=> r₃ = 9x

using this in (i), 9x/Î» = 3/2Î

=> Î» = 6xÎ

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$ourav @@@ -- WILL Never give ·2009-09-26 22:38:54

16. answer is 27........

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Asish Mahapatra ·2009-09-26 22:40:08

18. shortest wavelength occurs when transition occurs from âˆž

So, shortest lyman wavelength 1Î» = R

=> x = 1/R ... (i)

Longest wavelength occurs when transition takes place frm just above excited state to the prev, state,

So, longest Balmer wavelength for He⁺
1Î» = 4R(12²-13²)

=> 1/Î» = 5R/9

=> Î» = 9/5R = 9x/5

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nix_13 12 ·2009-09-26 23:00:27

maths_physco how??

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$ourav @@@ -- WILL Never give ·2009-09-26 23:03:25

c there are 9 orbitals to be filled up in period 4....so,acc to d ques each orbital can hold 3 electrond ...so number of elements are 3*9=27

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