General Knowledge - Chemistry - Discussion
Discussion Forum : Chemistry - Section 1 (Q.No. 51)
51.
The ionisation energy of hydrogen atom in the ground state is x KJ. The energy required for an electron to jump from 2nd orbit to 3rd orbit is
Discussion:
16 comments Page 2 of 2.
Rakesh said:
9 years ago
I can't understand this.
Kani said:
7 years ago
Can anyone explain this?
Nisarga ram said:
6 years ago
I can't understand please can anyone explain this.
Ramesh v said:
6 years ago
we know that E=X(1/M^2*1/N^2), where M=2,N=3 putting this value of M,N so,
E=x(1/2*1/3^2)
E=x(1/4-1/9).
E=x (9-4)/36.
E =x.5/36=5x/36 answer.
E=x(1/2*1/3^2)
E=x(1/4-1/9).
E=x (9-4)/36.
E =x.5/36=5x/36 answer.
(2)
Godfred said:
3 years ago
I'm confused. Please explain in more detail.
Khemi said:
2 months ago
E1 = -x
(energy of 1st orbit relative to free electron at 0).
For a hydrogen atom:
Energies of 2nd and 3rd orbit
E2 =(-x/2²) = (-x/4).
E3 = (-x/3²) = (-x/9).
Now,
Energy required to jump from 2 → 3.
∆ E = E3 - E2.
={-(x/9)- (-x/4)}.
= (X/4) - (x/9).
= (9x-4x)/36.
= 5x/36 Answer.
(energy of 1st orbit relative to free electron at 0).
For a hydrogen atom:
Energies of 2nd and 3rd orbit
E2 =(-x/2²) = (-x/4).
E3 = (-x/3²) = (-x/9).
Now,
Energy required to jump from 2 → 3.
∆ E = E3 - E2.
={-(x/9)- (-x/4)}.
= (X/4) - (x/9).
= (9x-4x)/36.
= 5x/36 Answer.
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