Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 40 - All About Atoms - Problems - Page 1247: 11

Answer

The magnitude of the orbital angular momentum$(\vec L)$ of an electron trapped in an atom is given by $L=\sqrt {l(l+1)}\hbar,\;\;\;\text{for}\;l=0,1,2,...,(n-1)$ The component $L_z$ of $\vec L$ on a $z$ axis is given by $L_z=m_l\hbar,\;\;\;\text{for}\;m_l=0,±1,±2,...,±l$ If, $L_x$, $L_y$ are the components of $\vec L$ along $x$ and $y$ axes, then we can write $L^2=L_x^2+L_y^2+L_z^2$ or, $L_x^2+L_y^2=L^2-L_z^2$ or, $L_x^2+L_y^2=(\sqrt {l(l+1)}\hbar)^2-(m_l\hbar)^2$ or, $L_x^2+L_y^2=[l(l+1)-m_l^2]\hbar^2$ or, $(L_x^2+L_y^2)^{1/2}=[l(l+1)-m_l^2]^{1/2}\hbar$ (proved)

Work Step by Step

The magnitude of the orbital angular momentum$(\vec L)$ of an electron trapped in an atom is given by $L=\sqrt {l(l+1)}\hbar,\;\;\;\text{for}\;l=0,1,2,...,(n-1)$ The component $L_z$ of $\vec L$ on a $z$ axis is given by $L_z=m_l\hbar,\;\;\;\text{for}\;m_l=0,±1,±2,...,±l$ If, $L_x$, $L_y$ are the components of $\vec L$ along $x$ and $y$ axes, then we can write $L^2=L_x^2+L_y^2+L_z^2$ or, $L_x^2+L_y^2=L^2-L_z^2$ or, $L_x^2+L_y^2=(\sqrt {l(l+1)}\hbar)^2-(m_l\hbar)^2$ or, $L_x^2+L_y^2=[l(l+1)-m_l^2]\hbar^2$ or, $(L_x^2+L_y^2)^{1/2}=[l(l+1)-m_l^2]^{1/2}\hbar$ (proved)
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