Question

A hydrogen atom has $l=2$. What are the (a) minimum (as a multiple of $h$ ) and (b) maximum values of the quantity ${\left(L_x^2+L_y^2\right)}^{1/2}$ ?

Short answer

Part (a) The minimum value of ${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{2}\hslash$

Part (b) The maximum value of ${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{6}\hslash$

Step-by-step solution

Step 1. Given information

The equation of total angular momentum,

$L^2=L_x^2+L_y^2+L_z^2$

Here,

$L=\sqrt{l(l+1)}\hslash$

$\left(L_x^2+L_y^2\right)=L^2-L_z^2$

Here $L_x,L_y,L_z$ are the components of angular momentum

${\left(L_z\right)}_{\max }=l\hslash$

${\left(L_z\right)}_{\min }=0$

Step 2. Finding the minimum value of Lx2+Ly212

$\left(L_x^2+L_y^2\right)=L^2-{\left(L_z\right)}_{\max }^2$

$=l(l+1){\hslash }^2-l^2{\hslash }^2$

${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{l}\hslash$

${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{2}\hslash$

The minimum value of${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{2}\hslash$

Step 3. Finding the maximum value of Lx2+Ly21/2=6ℏ

$\left(L_x^2+L_y^2\right)=L^2-{\left(L_z\right)}_{\min }^2$

$=l(l+1){\hslash }^2-0$

$\left(L_x^2+L_y^2\right)=l(l+1){\hslash }^2$

${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{l(l+1)}\hslash$

${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{2(2+1)}\hslash$

$=\sqrt{6}\hslash$

The maximum value of${\left(L_x^2+L_y^2\right)}^{1/2}=\sqrt{6}\hslash$