| List -I | List -II |
| (A) \[mvr=\frac{nh}{2\pi }\] | (i) Paschen series |
| (B) Infra - red | (ii) Electron total energy |
| (C) \[\lambda =\frac{h}{p}\] | (iii) de - Broglie equation |
| (D) \[\frac{-{{e}^{2}}}{2r}\] | (iv) Schrodinger equation |
| (v) Bohrs equation |
A) A-(v) B-(ii) C-(iii) D-(i)
B) A-(iii) B-(ii) C-(v) D-(iv)
C) A-(v) B-(i) C-(iii) D-(ii)
D) A-(iv) B-(i) C-(ii) D-(iii)
Correct Answer: C
Solution :
According of Bohr among the infinite number of possible circular orbitals, an electron can revolve only in those orbitals whose angular momentum (mvr) is an integral multiple of the factor \[h/2\pi \]ie, \[mvr=\frac{h}{2\pi}\] Hence, this equation is called Bohrs equation. In infra-red region, Paschen series of lines is obtained. \[\lambda =\frac{h}{P}\]is known as de-Broglie equation. Total energy of an electron \[\text{= KE + PE}\] \[=\frac{{{e}^{2}}}{2r}+\left( -\frac{{{e}^{2}}}{r} \right)\] \[=-\frac{{{e}^{2}}}{2r}\] Hence, \[A-(v),B-(i),C(iii),D-(ii)\]
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