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:
15 comments Page 1 of 2.
Rohit chauhan said:
1 decade ago
Could anybody explain me this question?
Kirtijain said:
1 decade ago
Please explain how this calculation is done ?
Vishnu said:
1 decade ago
Energy of electron transition depends on the principle quantum numbers.
ie: 1/n1square - 1/n2square.
Here n1=2, n2 =3.
ie: 1/n1square - 1/n2square.
Here n1=2, n2 =3.
(1)
Clint said:
1 decade ago
From Rydberg's:
x{[1/(2^2)] - [1/(3^2)]} = 5x/36.
x{[1/(2^2)] - [1/(3^2)]} = 5x/36.
Dania said:
1 decade ago
Can't understand this?
Nasim said:
10 years ago
Formula (1/n1^2-1/n2^2). In this cause n1 = 2 & n2 = 3. So 5x/36.
Sangeet sarkar said:
9 years ago
Couldn't get it, please explain breifly.
Uforo said:
9 years ago
Thanks, I understand the solution.
(1)
Vinay said:
9 years ago
I can not understand properly. Please explain in detail.
Jeffrey said:
9 years ago
I didn't get this. Please explain me clearly.
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