Breathing Fire

Chemistry Level 2

$\Large \Psi_{2s} = \frac{1}{4 \sqrt{2 \pi}} \left(\frac{1}{a_0}\right)^{\frac{3}{2}} \left(2- \frac{r}{a_0}\right) e^{\frac{-r}{a_0}}$

The Schrödinger's wave equation for a hydrogen atom is given above, where $a_0$ is the Bohr's radius of $0.529\ \overset{\circ}{\text{A}}$ . If the radial node in $2s$ is at $r_0$ , then what is the value of $r_0$ ?

Give your answer in Angstrom units: $1\ \overset{\circ}{\text{A}} = 10^{-10} \text{ m}$ .

The answer is 1.058.

2 solutions

Samarpit Swain
Jun 2, 2015

We can define the probability of finding an electron in the nodal plane for the $2s$ orbital as: $P=\Psi_{2s}^{2}.4\pi r^{2}dr=0$ . Therefore, $\frac{1}{32\pi} \left(\frac{1}{a_0}\right)^{3} \left(2- \frac{r}{a_0}\right)^{2} e^{\frac{-2r}{a_0}}\left(4\pi r^{2}dr\right)=0$ . From here we can collect the terms and check which of them will be feasible for satisfying the above equation.

Clearly $\left(2-\dfrac{r}{a_{0}}\right)^{2}$ has to be equal to $0$

=> $2-\dfrac{r}{a_{0}}=0$

$\therefore r_{0}= r= 2a_{0}=1.058Å :)$

Why did u multiplied it with 4πr^2dr

Akarsh Kumar Srit - 5 years, 5 months ago

This is the differential volume $dV$ . The Probability of finding an electron in a very small volume , according to the wave equation is defined as $P=\psi^2dV$

Samarpit Swain - 5 years, 5 months ago

Satvik Choudhary
Jun 3, 2015

Its just an old iit problem

@satvik choudhary I didn't know about that!

Ameya Salankar - 6 years ago

Yes it is an old iit problem....iit has asked Schrödinger's Wave Equation only once or twice

Samarth Agarwal - 5 years, 12 months ago

Breathing Fire

The answer is 1.058.

2 solutions

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