For an electron, with n=3 has only one radial node. If the orbital angular momentum of the electron is given as √ X (h/2 π), then value of X will be

Question

For an electron, with n=3 has only one radial node. If the orbital angular momentum of the electron is given as √ X (h/2 π), then value of X will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of radial nodes =n-l-1=1 =3-l-1=1 ∴ l=1 Orbital angular momentum =√(l+1) (h/2 π)=√2 (h/2 π)

Q. For an electron, with $n = 3$ has only one radial node. If the orbital angular momentum of the electron is given as $X \frac{h}{2 π}$ , then value of $X$ will be

Number of radial nodes $= n - l - 1 = 1$
$= 3 - l - 1 = 1$
$∴ l = 1$
Orbital angular momentum $= (l + 1) \frac{h}{2 π} = 2 \frac{h}{2 π}$

Q. For an electron, with n=3 has only one radial node. If the orbital angular momentum of the electron is given as X​2πh​, then value of X will be

Number of radial nodes =n−l−1=1 =3−l−1=1 ∴l=1 Orbital angular momentum =(l+1)​2πh​=2​2πh​

Q. For an electron, with $n = 3$ has only one radial node. If the orbital angular momentum of the electron is given as $X \frac{h}{2 π}$ , then value of $X$ will be

Number of radial nodes $= n - l - 1 = 1$
$= 3 - l - 1 = 1$
$∴ l = 1$
Orbital angular momentum $= (l + 1) \frac{h}{2 π} = 2 \frac{h}{2 π}$