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.
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)
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)
Nasim said:
10 years ago
Formula (1/n1^2-1/n2^2). In this cause n1 = 2 & n2 = 3. So 5x/36.
Vinay said:
9 years ago
I can not understand properly. Please explain in detail.
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.
Nisarga ram said:
6 years ago
I can't understand please can anyone explain this.
Kirtijain said:
1 decade ago
Please explain how this calculation is done ?
Jeffrey said:
9 years ago
I didn't get this. Please explain me clearly.
Godfred said:
3 years ago
I'm confused. Please explain in more detail.
Sangeet sarkar said:
9 years ago
Couldn't get it, please explain breifly.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers