91Ó°ÊÓ

The state is given as(n,l, m)=(2,1,0).

Orbitals are the regions in the space where electrons are found, and there is a very high probability of the presence of electrons in its orbital. The shape of the orbital is defined by the Azimuthal quantum number ‘l’.

To find the probability between $\mathit{θ}\mathbf{=}\mathbf{0}\mathbf{°}$and $\mathit{θ}\mathbf{=}\mathbf{60}\mathbf{°}$and between $\mathit{θ}\mathbf{=}\mathbf{60}\mathbf{°}$and $\mathit{θ}\mathbf{=}\mathbf{180}\mathbf{°}$. Due to symmetry, you will double the integral from $\mathit{θ}\mathbf{=}\mathbf{0}\mathbf{°}$to $\mathit{θ}\mathbf{=}\mathbf{60}\mathbf{°}$and will get our answer.

Where, $\mathit{θ}\mathbf{=}$Angle between electron and z-axis with respect to the origin.

Probability can be calculated as:

$$\begin{gathered} P_1=2{∫}_0^{\mathrm{π}/3}{\left(\sqrt{\frac{3}{4\mathrm{π}}}\cos \mathrm{θ}\right)}^{2}2\mathrm{π}\sin \mathrm{θ}d\mathrm{θ} \\ =3{∫}_0^{\mathrm{π}/3}{\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ} \\ =3{\overline{)\frac{-3{\cos }^3\mathrm{θ}}{3}}}_0^{\mathrm{π}/3} \\ =0.875 \end{gathered}$$

Thus, The probability that it would be found within 60 degrees of z-axis = 0.875.

Where, a₀= radius of the hydrogen atom

If only the radial part of the wave function is involved, and R_2,1(r) is the same for the (2,1,0) state as for a (2,1,+1) state,

Hence, Probability can be calculated as:

$$\begin{gathered} P_2={∫}_{\mathrm{π}/3}^{2\mathrm{π}/3}{\left(\sqrt{\frac{3}{8\mathrm{π}}}\sin \mathrm{θ}\right)}^{2}2\mathrm{π}\sin \mathrm{θ}d\mathrm{θ} \\ =\frac{3}{4}{∫}_{\mathrm{π}/3}^{2\mathrm{π}/3}{\sin }^3\mathrm{θ}d\mathrm{θ} \\ =\frac{3}{4}\frac{11}{12} \\ =0.688 \end{gathered}$$

Thus,the probability that it would be found between r = 2a₀ and r = 6a₀= 0.662.

Probability can be calculated as:

Probability = p1 x p2

= 0.875 x 0.662 [from eq. 1 and eq. 2]

= 0.58

Thus, the probability that it would be found within 60 degrees of the z-axis and between r = 2a₀ and r = 6a₀ = 0.58.