Part C: Concept Checks
Open the lesson
Quick concept checks for JEE atomic structure, taught like a live class: how to crack assertion–reason and true-or-false questions, why cathode rays have the same e/m in every gas, why 3s and 3d have equal energy in hydrogen, Bohr's model for He+, the photoelectric traps on intensity and frequency, the de Broglie wavelength of a cricket ball, and why the ionisation energy of He+ is four times that of hydrogen.
Chapters
- 0:00Introduction
- 0:45Assertion–Reason: ask three questions
- 1:41One electron: the energy depends only on n
- 3:04True or false: the photoelectric effect
- 4:30Example 1: de Broglie wavelength of a cricket ball
- 5:33Example 2: ionisation energy of He+ is four times H
- 7:19Quick revision: three questions, Z squared, photocurrent
Transcript
0:00Introduction
Hello everyone! Here's one quick question before we start. Hydrogen has one proton in its nucleus. The helium ion, H e plus, has two protons, but, just like hydrogen, only one electron. So, how much more energy does it take to pull that one electron out? Two protons, so twice the energy, right? Hmm, are you sure? Keep your answer ready; example two will settle it. Today is a concept check class: assertion and reason, and true or false, from the whole of atomic structure. And look, in JEE, these short questions look easy, but this is exactly where good students lose marks. So, let's begin.
0:45Assertion–Reason: ask three questions
So, first, how do we attack an assertion and reason question? Here's the method: ask three things, in order. One: is the assertion, the capital A on screen, true? Two: is the reason, the R, true? And three: does the reason explain the assertion? One question at a time. Okay? Let's try one. Assertion: cathode rays have the same e by m, whatever the gas in the tube. Here, e by m means the charge of the particle divided by its mass, okay? Reason: cathode rays are streams of electrons. Now you tell me: are both true, and does the reason explain the assertion? Both true, yes, and the reason explains it! Cathode rays are electrons, and electrons are the same in every material. Change the gas, and you still get the same electrons, with the same e by m. Really simple, right?
1:41One electron: the energy depends only on n
Next one. Assertion: in hydrogen, a three s electron and a three d electron have equal energy. Hmm. But you learnt that three s fills long before three d, didn't you? So three s must be lower, and the assertion is false, right? No, no, careful! Read it again: in hydrogen. And here is the reason: in a one-electron species, the energy depends only on n, the principal quantum number. Three s and three d both have n equal to three, so they have the same energy. And why only n? With just one electron, there is no electron–electron repulsion to split the subshells. That splitting, and the filling order you learnt, belong to many-electron atoms. So, both true, and the reason explains. Clear? Now a twist. Assertion: Bohr's model explains the spectrum of H e plus, the helium ion; the small plus up top means it has lost one electron. Reason: H e plus has one electron. But Bohr's model fails for helium, doesn't it? So the assertion is false? Wait, read carefully! The model fails for the helium atom, because its two electrons repel each other. But H e plus has just one electron left, like hydrogen. And Bohr's model applies to any one-electron species. So, both true, and the reason explains. Absolutely!
3:04True or false: the photoelectric effect
Now, true or false, on the photoelectric effect. Read them with me. One: the photocurrent increases with the frequency of the light. Two: the stopping potential does not depend on the intensity. Three: no electrons are emitted below the threshold frequency, however intense the light. Statement one first. Higher frequency means more energetic photons, so more current, right? No, no, careful! Look at the right graph. I is the photocurrent, and V is the voltage. At a fixed frequency, the bright beam gives a bigger current than the dim one. So the current depends on intensity, not frequency. Statement one is false. Statement two. Both curves fall to zero at the same point, minus V naught; V naught, V with a small zero, is the stopping potential. Bright or dim, the same V naught, right? So, true. Statement three. On the left graph, K max, the largest kinetic energy of the electrons, is plotted against nu, the frequency. The blue line, for metal one, starts only at nu naught, its threshold frequency; metal two starts later, at its own threshold. Below the threshold, no electrons, however intense the light. True! So the answer is: false, true, true. Look, intensity decides how many electrons come out; frequency decides whether they come out at all. Remember this line.
4:30Example 1: de Broglie wavelength of a cricket ball
Now, example one. Assertion: the de Broglie wavelength of a cricket ball is never observed. Reason: it is far smaller than any object or opening. Use our three questions. Pause the video, and try it first. Done? Okay, let's check. The de Broglie wavelength: lambda, this Greek letter, equals h over m v. That is Planck's constant, divided by the mass times the speed. Now, for a heavy cricket ball, how big will this lambda be? Tiny, yes! About ten to the power minus thirty four metre; the wavy equals sign means approximately. Planck's constant is so small that, for anything as heavy as a ball, lambda becomes tiny. That is far smaller than any object or opening, so no wave effects can ever be seen. So the assertion is true, the reason is true, and the reason is exactly why. But notice: the ball does have a wavelength; we just cannot see it. Got it?
5:33Example 2: ionisation energy of He+ is four times H
Now the harder one, JEE level, and it is our opening puzzle. Assertion: the ionisation energy of H e plus is four times that of hydrogen. Reason: for one-electron species, the energy of a level is proportional to Z squared, where Z is the atomic number. Pause and try. Hint: both have one electron. Done? Alright, first the quick answer from the start. Helium has two protons, hydrogen has one. So twice the energy, and the assertion is false, right? No, no, careful! The energy goes as Z squared, not simply as Z, okay? From the ground state, the ionisation energy, the energy to pull the electron out, is thirteen point six times Z squared, in electron volts, the e V on screen. With me? Hydrogen, the H on screen: Z is one. Thirteen point six times one squared: thirteen point six electron volts. In the figure, hydrogen's lowest level, n equal to one, sits at minus thirteen point six: exactly that deep. Now H e plus: Z is two, and two squared is four. Thirteen point six times four: fifty four point four electron volts. Look at the orange ladder on the right: its lowest level sits all the way down at minus fifty four point four. And see the arrow marked equal: the second level of H e plus lies exactly at minus thirteen point six, level with the ground state of hydrogen. Interesting, isn't it? Fifty four point four divided by thirteen point six: exactly four. So the assertion is true, the reason is true, and Z squared is exactly why. Two protons, four times the energy. Our opening puzzle, solved!
7:19Quick revision: three questions, Z squared, photocurrent
So, quick revision. One: for assertion and reason, ask: is the assertion true, is the reason true, and does the reason explain it? Two: with one electron, the energy depends only on n, and the ionisation energy is thirteen point six Z squared electron volts. Three: the photocurrent is set by intensity, not frequency, and below nu naught, no electrons at all. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!
More videos in Unit 2 · Atomic Structure
- Discovery of Subatomic Particles
- Atomic Models
- Atomic Number, Mass Number and Related Terms
- Counting Particles and Average Atomic Mass
- Wave Nature of Radiation
- Planck's Quantum Theory
- Counting Photons and Energy Balance
- The Photoelectric Effect
- Bohr's Postulates and the Orbits
- Transitions, Ionisation and Excitation
- Other Orbital Quantities and Limitations
- Hydrogen-like Ions: A Drill
- Line Spectra and the Rydberg Formula
- First Lines, Series Limits and Ratios
- How Many Lines?
- de Broglie Wavelength
- de Broglie Waves in Bohr Orbits
- Heisenberg's Uncertainty Principle
- Wave Functions and Orbitals
- The Four Quantum Numbers
- Shapes of Orbitals and Nodes
- Energies of Orbitals
- Rules for Filling Orbitals
- Exceptions and Exchange Energy
- Configurations of Ions
- Unpaired Electrons and Magnetic Moment
- Counting Electrons by Quantum Numbers
- Overview
- Mixed Problems: Set 1 (Particles, Photons and the Photoelectric Effect)
- Mixed Problems: Set 2 (Bohr Model, Spectra and de Broglie)
- Mixed Problems: Set 3 (Quantum Numbers, Nodes and Configurations)
- Overview
- Part A: JEE Main Level