Atomic StructureEasy
Question
The electron in a hydrogen atoms makes a transition n1 → n2, n1 and n2 are the principal quantum number of two states. Assume the Bohr′s model to be valid. The time period of the electrons in the initial state is eight times that in the final state. The possible values of n1 and n2 are
Options
A.n1 = 4, n2 = 2
B.n1 = 8, n2 = 2
C.n1 = 8, n2 = 1
D.n1 = 6, n2 = 3
Solution
Time period, Tn = 
i.e., Tn ∝
But rn ∝ n2 and
Therefore, Tn ∝ n3
Given, Tn1 = 8Tn2
Hence, n1 = 2n2
Therefore, options (a) and (d) are correct.
i.e., Tn ∝
But rn ∝ n2 and
Therefore, Tn ∝ n3
Given, Tn1 = 8Tn2
Hence, n1 = 2n2
Therefore, options (a) and (d) are correct.
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The smallest wavelength of Lyman series is 91 nm . The difference between the largest wavelengths of Paschen and Balmer ...When a hydrogen atom, initially at rest emits a photon resulting in transition n = 5 → n = 1. its recoil speed is ...The different lines in the Lyman series have their wavelengths lying between :...Starting with a simple of pure 66Cu, of it decays in to Zn in 15 minutes. The corresponding half life is...In Bohr′s atom the number of de broglie waves associated with electron moving in the permitted orbit is -...