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Electron Distribution in Orbits

arrange the e...

Question

arrange the electrons represented by the following sets of quantum nos. in the decreasing order of energy for a multi electron atom: (1) n=4,l=0,m=0,s=+1\2, (2) n=3, l=1, m=1, s=-1\2 (3) n=3, l=2, m=0, s=+1\2 (4) n=3, l=0, m=0, s=-1\2

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Solution

$T h e v a r i o u s o r b i t a l s a r e (1) n = 4, l = 0, m = 0, s = + 1 / 2 r e p r e s e n t s 4 s (2) n = 3, l = 1, m = 1, s = - 1 / 2 r e p r e s e n t s 3 p (3) n = 3, l = 2, m = 0, s = + 1 / 2 r e p r e s e n t s 3 d (4) n = 3, l = 0, m = 0, s = - 1 / 2 r e p r e s e n t s 3 s (a l l t h e s e o r b i t a l s a r e a s s i g n e d b a s e d o n t h e n v a l u e a s t h e s t a r t i n g n u m b e r a n d l d e c i d e s t h e o r b i t a l . T h a t i s l = 0 m e a n s s o r b i t a l, l = 1 m e a n s p o r b i t a l, l = 2 m e a n s d o r b i t a l, l = 3 m e a n s f o r b i t a l) A c c o r d i n g t o A u f b a u p r i n c i p l e, t h e o r d e r o f d e c r e a \sin g e n e r g y i s 3 d > 4 s > 3 p > 3 s$

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Similar questions

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

n = 4, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 1, m_{l} = 1, m_{s} = - \frac{1}{2}

n = 3, l = 2, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 0, m_{l} = 0, m_{s} = - \frac{1}{2}

\pm

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 2, m = + 1, s = - \frac{1}{2}

n = 3, l = 2, m = - 1, s = + \frac{1}{2}

n = 3, l = 1, m = 0, s = + \frac{1}{2}

n = 5, l = 2, m = + 1, s = - \frac{1}{2}

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