Question

An electron in a hydrogen atom is in the (n,l,m_l) = (2,1,0) state.

(a) Calculate the probability that it would be found within 60 degrees of z-axis, irrespective of radius.

(b) Calculate the probability that it would be found between r = 2a₀ and r = 6a₀, irrespective of angle.

(c) What is the probability that it would be found within 60 degrees of the z-axis and between r = 2a₀ and r = 6a₀?

Short answer

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

(b) The probability that it would be found between r = 2a₀ and r = 6a₀= 0.662.

(c) The probability that it would be found within 60 degrees of the z-axis and between r = 2a₀ and r = 6a₀ = 0.58.

Step-by-step solution

Given data

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

(a) Probability that electron would be found within 60 degrees of the z-axis

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

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

Probability can be calculated as:

$$\begin{gathered} P_1=2{\int }_0^{\pi /3}{\left(\sqrt{\frac{3}{4\pi }}\cos \theta \right)}^{2}2\pi \sin \theta d\theta \\ =3{\int }_0^{\pi /3}{\cos }^2\theta \sin \theta d\theta \\ =3{\overline{)\frac{-3{\cos }^3\theta }{3}}}_0^{\pi /3} \\ =0.875 \end{gathered}$$

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

(b) Probability that it would be found between r = 2a0 and r = 6a0

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:

$$\begin{gathered} P_2={\int }_{\pi /3}^{2\pi /3}{\left(\sqrt{\frac{3}{8\pi }}\sin \theta \right)}^{2}2\pi \sin \theta d\theta \\ =\frac{3}{4}{\int }_{\pi /3}^{2\pi /3}{\sin }^3\theta d\theta \\ =\frac{3}{4}\frac{11}{12} \\ =0.688 \end{gathered}$$

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

(c) The probability that it would be found within 60 degrees of the z-axis and between r = 2a0 and r = 6a0

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.