Answer this chemistry question from atomc structure!
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The formula for number of spectral lines is
No.of lines= n(n-1)/2
So,
10= n^2 - n /2
n^2 - n = 20
n^2 -n-20=0
(n-5)(n+4) = 0
Therefore, the principal quantum number cannot be negative and hence the n = 5
Now, This is thus Lymann transfer because at last it's coming to the ground state,
and we know that lines becomes visible only in the Balmer and Paschen transition.
Thus, The spectral lines which will be visible are
(3,1)(3,2)(2,1).
Hence, the correct answer is (3)
No.of lines= n(n-1)/2
So,
10= n^2 - n /2
n^2 - n = 20
n^2 -n-20=0
(n-5)(n+4) = 0
Therefore, the principal quantum number cannot be negative and hence the n = 5
Now, This is thus Lymann transfer because at last it's coming to the ground state,
and we know that lines becomes visible only in the Balmer and Paschen transition.
Thus, The spectral lines which will be visible are
(3,1)(3,2)(2,1).
Hence, the correct answer is (3)
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