| Column I | Column II | ||
| A. | \[mvr=\frac{nh}{2\pi }\] | 1. | De-Broglie equation |
| B. | \[\lambda =\frac{h}{p}\] | 2. | Electrons total energy |
| C. | \[-\frac{{{e}^{2}}}{2r}\] | 3. | Paschen series |
| D. | Infrared | 4. | Bohrs equation |
A) A-1 B-3 C-4 D-2
B) A-4 B-1 C-2 D-3
C) A-4 B-1 C-3 D-2
D) A-2 B-1 C-3 D-4
E) A-1 B-2 C-3 D-4
Correct Answer: B
Solution :
Option is correct. According to Bohr, among the infinite number of possible circular orbitals, an electron can remove only in those orbitals whose angular momentum (mvr) is an integral multiple of the factor\[\frac{h}{2\pi }.\] \[mvr=\frac{nh}{2\pi }\] Total energy \[\Rightarrow \]\[E=KE+PE\] \[=\frac{{{e}^{2}}}{2r}+\left( \frac{-\,{{e}^{2}}}{r} \right)=\frac{-\,{{e}^{2}}}{2r}\] \[\lambda =\frac{h}{p}\xrightarrow{{}}\]de-Broglie equation In infrared region, Paschen series of lines is obtained.
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