schodinger wave equation jee

Prompt

The important output is: \boxed{\psi_{n,l,m_l}} $$ and from that you extract: * orbital * probability density * nodes * orbital shape * orientation * radial distribution * quantum numbers This is exactly why JEE questions can throw a disgusting-looking expression like the ones in your image and still expect you to solve it in 20–30 seconds. --- # 2. What the hell is \(\psi\)? This is the most important concept. $$ \boxed{\psi = \text{wavefunction}} $$ But **\(\psi\) itself isn't probability.** Instead: $$ \boxed{|\psi|^2=\text{probability density}} $$ Think of \(\psi\) as the **amplitude of an electron wave**. Then: $$ |\psi|^2 $$ tells you how strongly the electron is concentrated there. ### Analogy Imagine a sound wave. The amplitude can be positive or negative. But the **intensity** is related to amplitude squared, so it can't be negative. Same idea: $$ \psi \rightarrow \text{amplitude} $$ $$ |\psi|^2\rightarrow \text{probability density} $$ Therefore: $$ \boxed{\psi=0\Rightarrow |\psi|^2=0} $$ That gives us the whole concept of **nodes**. --- # 3. NODE = place where \(\psi=0\) This is one of the biggest JEE shortcuts. At a node: $$ \boxed{\psi=0} $$ therefore $$ \boxed{|\psi|^2=0} $$ So there is zero probability density there. There are two major types: ### Radial node A spherical surface around the nucleus. Example: $$ r=2a_0 $$ means: > At a spherical shell of radius \(2a_0\), \(\psi=0\). ### Angular node A plane/cone passing through the nucleus where the angular part becomes zero. Example: $$ \cos\theta=0 $$ gives: $$ \theta=90^\circ $$ That's the **\(xy\)-plane**. So: $$ \boxed{\text{Angular node = direction where }\psi=0} $$ --- # 4. The wavefunction has TWO major parts This is the key to understanding the expressions in your photo. For hydrogen-like atoms: $$ \boxed{\psi_{nlm}=R_{nl}(r)\,Y_l^{m}(\theta,\phi)} $$ Meaning: $$ \boxed{\psi=\text{RADIAL part}\times\text{ANGULAR part}} $$ Think: | Part | Depends on | Controls | | ------------------ | -------------- | ------------------------------- | | \(R(r)\) | distance \(r\) | radial behaviour/nodes | | \(Y(\theta,\phi)\) | direction | shape/orientation/angular nodes | This one distinction solves a **huge fraction** of Schrödinger questions. --- # 5. Your first wavefunction in the image You have something like: $$ \psi \propto \left(\frac Za_0\right)^{3/2} e^{-Zr/a_0} $$ Notice: **Only \(r\)** appears. There is no \(\theta\), no \(\phi\). Therefore the wavefunction is spherically symmetric. That screams: $$ \boxed{s\text{-orbital}} $$ For example: $$ \boxed{1s} $$ has: $$ \psi_{1s}\propto \left(\frac Za_0\right)^{3/2} e^{-Zr/a_0} $$ Since exponential is never zero: $$ e^{-Zr/a_0}\neq0 $$ and the prefactor isn't zero. Therefore: $$ \boxed{\text{1s has 0 nodes}} $$ Makes sense from the formula: $$ \text{total nodes}=n-1 $$ For \(1s\): $$ n=1 $$ so: $$ 1-1=0 $$ --- # 6. Your second expression You have: $$ \psi\propto \left(\frac Za_0\right)^{3/2} r e^{-Zr/(2a_0)} \cos\theta $$ Now split it: ### Radial: $$ r e^{-Zr/(2a_0)} $$ ### Angular: $$ \cos\theta $$ The \(\cos\theta\) immediately tells you: $$ \boxed{p_z\text{-type}} $$ because $$ \boxed{p_z\propto\cos\theta} $$ And the \(r\) multiplying the exponential is characteristic of a \(2p\)-type radial function. So this is: $$ \boxed{2p_z} $$ --- # 7. Now look for nodes This is where JEE starts becoming fun. We have: $$ \psi\propto r e^{-Zr/(2a_0)}\cos\theta $$ For \(\psi=0\): ### Possibility 1: $$ r=0 $$ That's the origin, but **don't automatically count this as a radial node.** ### Possibility 2: $$ \cos\theta=0 $$ Therefore: $$ \theta=90^\circ $$ That's an angular node. So: $$ \boxed{\text{angular nodes}=1} $$ And for \(2p\): $$ n=2,\quad l=1 $$ Radial nodes: $$ n-l-1 $$ $$ =2-1-1=0 $$ Thus: $$ \boxed{2p:\quad 0\text{ radial},\ 1\text{ angular},\ 1\text{ total}} $$ --- # 8. WHY is it called \(p_z\)? This is the intuition part. Take: $$ \psi\propto\cos\theta $$ At the \(+z\) direction: $$ \theta=0^\circ $$ so: $$ \cos0=1 $$ Maximum positive amplitude. At the \(-z\) direction: $$ \theta=180^\circ $$ $$ \cos180=-1 $$ So amplitude has opposite sign. At the \(xy\)-plane: $$ \theta=90^\circ $$ $$ \cos90=0 $$ So: $$ \boxed{xy\text{ plane is the node}} $$ That's why \(p_z\) looks like two lobes on opposite sides of the \(xy\)-plane. **Important:** the \(+\) and \(-\) signs of \(\psi\) are NOT positive/negative probability. Because: $$ |\psi|^2 $$ is always positive. The signs represent **phase** of the wavefunction. --- # 9. Your third expression is even more important You have approximately: $$ \psi\propto \left(\frac Za_0\right)^{3/2} \left(\frac{Zr}{a_0}\right)^2 e^{-Zr/(3a_0)} (3\cos^2\theta-1) $$ Now split it. ### Radial: $$ \left(\frac{Zr}{a_0}\right)^2 e^{-Zr/(3a_0)} $$ ### Angular: $$ 3\cos^2\theta-1 $$ The angular factor: $$ \boxed{3\cos^2\theta-1} $$ is the famous \(d_{z^2}\)-type angular dependence. So this is: $$ \boxed{3d_{z^2}} $$ --- # 10. How do you know it's 3d? Look at the radial part: $$ r^2e^{-Zr/(3a_0)} $$ The denominator \(3a_0\) in the exponential strongly points toward \(n=3\). And the angular factor corresponds to: $$ l=2 $$ Therefore: $$ \boxed{n=3,\quad l=2} $$ so: $$ \boxed{3d} $$ --- # 11. Nodes of 3d Use the master formulas: $$ \boxed{\text{Total nodes}=n-1} $$ $$ \boxed{\text{Angular nodes}=l} $$ $$ \boxed{\text{Radial nodes}=n-l-1} $$ For \(3d\): $$ n=3,\quad l=2 $$ Therefore: ### Total $$ 3-1=2 $$ ### Angular $$ 2 $$ ### Radial $$ 3-2-1=0 $$ Thus: $$ \boxed{3d:\ 2\text{ angular nodes},\ 0\text{ radial nodes}} $$ --- # 12. BUT there's a beautiful shortcut here Suppose the question gives: $$ \psi\propto (3\cos^2\theta-1) $$ and asks for angular nodes. Set: $$ \psi=0 $$ So: $$ 3\cos^2\theta-1=0 $$ $$ \cos^2\theta=\frac13 $$ $$ \cos\theta=\pm\frac1{\sqrt3} $$ There are **two cones**. Those are the two angular nodal surfaces of \(d_{z^2}\). So you don't even need to remember a picture. --- # 13. The MASTER node formula This should be permanently in your notebook: $$ \boxed{\begin{aligned} \text{Total nodes}&=n-1\\ \text{Angular nodes}&=l\\ \text{Radial nodes}&=n-l-1 \end{aligned}} $$ Examples: | Orbital | \(n\) | \(l\) | Radial | Angular | Total | | ------- | ----: | ----: | -----: | ------: | ----: | | 1s | 1 | 0 | 0 | 0 | 0 | | 2s | 2 | 0 | 1 | 0 | 1 | | 2p | 2 | 1 | 0 | 1 | 1 | | 3s | 3 | 0 | 2 | 0 | 2 | | 3p | 3 | 1 | 1 | 1 | 2 | | 3d | 3 | 2 | 0 | 2 | 2 | | 4s | 4 | 0 | 3 | 0 | 3 | | 4p | 4 | 1 | 2 | 1 | 3 | | 4d | 4 | 2 | 1 | 2 | 3 | | 4f | 4 | 3 | 0 | 3 | 3 | Notice something beautiful: For a fixed \(n\): $$ \boxed{\text{all orbitals have }n-1\text{ total nodes}} $$ --- # 14. Why does \(s,p,d,f\) correspond to \(l=0,1,2,3\)? Another absolute JEE must-know: $$ \boxed{ \begin{array}{c|c} l&\text{subshell}\\ \hline 0&s\\ 1&p\\ 2&d\\ 3&f \end{array}} $$ Therefore: $$ l=0\Rightarrow s $$ $$ l=1\Rightarrow p $$ $$ l=2\Rightarrow d $$ $$ l=3\Rightarrow f $$ And because angular nodes \(=l\): $$ \boxed{s:0,\quad p:1,\quad d:2,\quad f:3} $$ --- # 15. What about \(m_l\)? This is where people often mix things up. Quantum numbers: $$ \boxed{n,\ l,\ m_l,\ m_s} $$ ### \(n\) Controls: * shell * approximate size * energy in hydrogen-like species * total nodes ### \(l\) Controls: * subshell * orbital shape * angular nodes ### \(m_l\) Controls: * orientation of the orbital in space $$ m_l=-l,\ldots,0,\ldots,+l $$ Therefore number of possible \(m_l\) values: $$ \boxed{2l+1} $$ So: ### s $$ l=0 $$ $$ m_l=0 $$ 1 orbital. ### p $$ l=1 $$ $$ m_l=-1,0,+1 $$ 3 orbitals. ### d $$ l=2 $$ $$ m_l=-2,-1,0,+1,+2 $$ 5 orbitals. --- # 16. One subtle Advanced-level point Don't blindly say: > \(m_l=0\) means \(p_z\). In the **real-orbital representation commonly used in chemistry**, \(p_z\) corresponds to \(m_l=0\), but \(p_x\) and \(p_y\) are real combinations of the \(m_l=\pm1\) states. For JEE chemistry, the useful practical picture is: $$ p_x,\ p_y,\ p_z $$ are three mutually perpendicular \(p\) orbitals. The important point is: $$ \boxed{m_l\text{ determines orientation information, not the energy for a hydrogen-like atom}} $$ --- # 17. VERY important: \(\psi^2\) vs radial probability This is probably the biggest conceptual trap in these questions. Suppose: $$ |\psi|^2 $$ is high at some point. That means **probability density at that point** is high. But suppose they ask: > Probability of finding electron between \(r\) and \(r+dr\). Now you need the volume of the spherical shell: $$ dV=4\pi r^2dr $$ for a spherically symmetric orbital. Therefore: $$ \boxed{dP=4\pi r^2|\psi|^2dr} $$ For a general hydrogen orbital after angular integration: $$ \boxed{P(r)\propto r^2|R(r)|^2} $$ This is called the **radial probability distribution**. --- # 18. Why does the \(r^2\) appear? This gives amazing intuition. Near the nucleus: $$ r\approx0 $$ The spherical shell has almost zero volume. Even if the probability density is large there, there isn't much space. As \(r\) increases, the shell gets bigger: $$ \text{shell area}\propto r^2 $$ Eventually the exponential decay of the wavefunction wins. So radial probability can have a maximum at some nonzero radius. That's why: $$ \boxed{\text{Most probable radius}\neq\text{point of maximum }|\psi|^2} $$ This distinction is VERY useful for Advanced. --- # 19. Example: 1s For hydrogen: $$ \psi_{1s}\propto e^{-r/a_0} $$ Therefore: $$ |\psi|^2\propto e^{-2r/a_0} $$ This is maximum at: $$ \boxed{r=0} $$ So **probability density** is maximum at the nucleus. But radial probability: $$ P(r)\propto r^2e^{-2r/a_0} $$ has its maximum at: $$ \boxed{r=a_0} $$ That seems contradictory until you remember: * density asks **how concentrated at a point** * radial probability asks **how much probability is contained in an entire spherical shell** --- # 20. How \(Z\) changes the wavefunction Your equations contain: $$ \frac Za_0 $$ This is important. For a hydrogen-like ion: $$ \boxed{\text{effective size}\sim\frac{a_0}{Z}} $$ Higher \(Z\): $$ \boxed{\text{electron cloud contracts}} $$ So: $$ Z\uparrow \Rightarrow r\downarrow $$ roughly. And energy: $$ \boxed{E_n=-13.6\frac{Z^2}{n^2}\text{ eV}} $$ So: $$ Z\uparrow\Rightarrow |E|\uparrow $$ Electron is more tightly bound. --- # 21. Scaling trick for hydrogen-like atoms Suppose you know some radius/probability feature for H. For hydrogen-like ion with nuclear charge \(Z\): $$ \boxed{r_Z=\frac{r_H}{Z}} $$ For example, if a radial node occurs at: $$ 2a_0 $$ for H, then for \(He^+\): $$ Z=2 $$ the corresponding distance becomes: $$ \boxed{a_0} $$ This is a very nice Advanced shortcut. --- # 22. How to attack a scary given wavefunction Suppose JEE gives: $$ \psi=C \left(\frac{Z}{a_0}\right)^{3/2} \left(\frac{Zr}{a_0}\right)^2 e^{-Zr/(3a_0)} (3\cos^2\theta-1) $$ DON'T stare at the entire monster. Use this algorithm: ### STEP 1 — Ignore constants Ignore: $$ C,\quad \left(\frac Z{a_0}\right)^{3/2} $$ They usually don't affect nodes/shape. ### STEP 2 — Split radial/angular $$ \boxed{\text{radial}\times\text{angular}} $$ Here: $$ R(r)\sim r^2e^{-Zr/(3a_0)} $$ $$ Y(\theta)\sim3\cos^2\theta-1 $$ ### STEP 3 — Identify \(l\) Angular part tells you shape. $$ 3\cos^2\theta-1 \Rightarrow d_{z^2} $$ Therefore: $$ l=2 $$ ### STEP 4 — Identify \(n\) Exponential/radial structure points toward \(n=3\). Thus: $$ n=3 $$ ### STEP 5 — Nodes $$ N_{\rm radial}=n-l-1 $$ $$ =3-2-1=0 $$ $$ N_{\rm angular}=l=2 $$ $$ N_{\rm total}=2 $$ DONE. --- # 23. The JEE question types you should be prepared for I've grouped the questions you should expect into **8 models**. ### MODEL 1 — Identify orbital from \(\psi\) Given: $$ \psi\propto re^{-r/2a_0}\cos\theta $$ You identify: $$ \boxed{2p_z} $$ --- ### MODEL 2 — Find nodes from given \(\psi\) Given: $$ \psi\propto (2-r/a_0)e^{-r/2a_0} $$ Node: $$ 2-r/a_0=0 $$ $$ \boxed{r=2a_0} $$ This exact style is a known JEE Advanced question. ([ExamSIDE][2]) --- ### MODEL 3 — Number of radial/angular nodes Given \(4d\): $$ n=4,\quad l=2 $$ Radial: $$ 4-2-1=1 $$ Angular: $$ 2 $$ Total: $$ 3 $$ --- ### MODEL 4 — Probability density Given: $$ \psi=A(2-r/a_0)e^{-r/2a_0} $$ Ask probability density. Immediately: $$ \boxed{|\psi|^2} $$ so square EVERYTHING: $$ |\psi|^2 =A^2(2-r/a_0)^2e^{-r/a_0} $$ --- ### MODEL 5 — Compare probability at two points If: $$ \psi_1=A $$ and: $$ \psi_2=2A $$ then: $$ \frac{P_2}{P_1} = \frac{|2A|^2}{|A|^2} = \boxed4 $$ **Amplitude doubles → probability density becomes 4×.** --- ### MODEL 6 — Radial probability Given \(R(r)\), don't simply calculate: $$ R^2 $$ You need: $$ \boxed{P(r)\propto r^2R^2(r)} $$ This is a favorite conceptual trap. --- ### MODEL 7 — Find angular nodal surface Given: $$ \psi\propto\cos\theta $$ Set: $$ \cos\theta=0 $$ $$ \theta=90^\circ $$ Therefore: $$ \boxed{xy\text{-plane}} $$ --- ### MODEL 8 — \(Z\)-dependence Given H and \(He^+\): $$ Z_H=1,\qquad Z_{He^+}=2 $$ Use: $$ r\propto\frac1Z $$ and: $$ E\propto-Z^2 $$ So \(He^+\) is: * smaller * more tightly bound * higher magnitude binding energy --- # 24. The intuition map I want you to remember Don't memorize Schrödinger as a pile of formulas. Think: $$ \boxed{ \text{Schrödinger equation} \rightarrow \psi \rightarrow |\psi|^2 \rightarrow \text{probability} } $$ Then: $$ \boxed{ \psi=R(r)Y(\theta,\phi) } $$ which splits into: $$ \boxed{ \begin{array}{ccc} R(r)&\longrightarrow&\text{radial behaviour}\\ Y(\theta,\phi)&\longrightarrow&\text{shape/orientation} \end{array}} $$ Then: $$ \boxed{ \begin{array}{ccc} n&\rightarrow&\text{shell/size/energy}\\ l&\rightarrow&\text{subshell/shape/angular nodes}\\ m_l&\rightarrow&\text{orientation}\\ m_s&\rightarrow&\text{spin} \end{array}} $$ And finally: $$ \boxed{ \begin{array}{ccc} \text{radial nodes}&=&n-l-1\\ \text{angular nodes}&=&l\\ \text{total nodes}&=&n-1 \end{array}} $$ That is the **core engine** behind this entire topic.