What is the smallest possible value for the angular momentum L of an electron in the n=5 shell?
Lmin= 0 kg*m^2/s
What is the largest possible value for the angular momentum L of an electron in the n=5 shell?
What is the Lmax? Please help I don't get this.Thank you!
L=ℏ √l(l+1) =√l(l+1)• h/2π
l= n-1 =>
l=0, 1, 2, 3,4.
L(max) = √(4•5) • 6.63•10⁻³⁴/2π=4.72•10⁻³⁴ kg•m²/s
L(min)= 0
To determine the largest possible value for the angular momentum (L) of an electron in the n=5 shell, we need to consider the values of the orbital angular momentum quantum number (l).
In the n=5 shell, the value of l can range from 0 to 4. The angular momentum (L) is given by the formula:
L = √(l(l+1) * ħ)
where ħ (h-bar) is the reduced Planck's constant.
To find the largest possible value for L, we need to find the maximum value for l within the n=5 shell. Since l can range from 0 to n-1, the maximum value for l in this case is 4.
Using the formula above, we can calculate L for l=4:
L = √(4(4+1) * ħ)
L = √(20 * ħ)
It is important to note that ħ has a value of 1.05457182 × 10^-34 kg*m²/s in SI units. Therefore, we can substitute this value into the equation to find the numerical value of L:
L = √(20 * 1.05457182 × 10^-34 kg*m²/s)
L ≈ 2.65 × 10^-34 kg*m²/s
Therefore, the largest possible value for the angular momentum (Lmax) of an electron in the n=5 shell is approximately 2.65 × 10^-34 kg*m²/s.