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JEE Main Physical Chemistry Flashcards

52 question-and-answer cards covering Physical Chemistry as it is examined in JEE Main. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.

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24 sample cards from the Physical Chemistry deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

  1. What constitutes electromagnetic radiation, and how do the electric and magnetic fields oscillate?

    EM radiation consists of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation. It is a transverse wave that travels through vacuum at the speed of light without needing a medium.

  2. Write the relationship between speed, frequency, and wavelength of EM radiation, and give the speed of light.

    $$c = \nu \lambda$$ where $c = 3 \times 10^{8}\ \text{m s}^{-1}$ is the speed of light, $\nu$ is frequency, and $\lambda$ is wavelength.

  3. Define wavenumber and give its formula and unit.

    Wavenumber $\bar{\nu}$ is the number of waves per unit length, the reciprocal of wavelength: $$\bar{\nu} = \frac{1}{\lambda}$$ Common unit: $\text{cm}^{-1}$ or $\text{m}^{-1}$.

  4. What is Planck's quantum theory, and write the energy of one quantum?

    Energy is emitted or absorbed not continuously but in discrete packets called quanta (photons). The energy of one quantum is $$E = h\nu = \frac{hc}{\lambda}$$ where $h = 6.626 \times 10^{-34}\ \text{J s}$ is Planck's constant.

  5. What is the photoelectric effect?

    The photoelectric effect is the ejection of electrons from a metal surface when light of suitable frequency strikes it. The ejected electrons are called photoelectrons.

  6. Define threshold frequency ($\nu_0$) and work function ($W_0$) in the photoelectric effect.

    Threshold frequency $\nu_0$ is the minimum frequency of incident light required to eject electrons from a metal. The work function $W_0 = h\nu_0$ is the minimum energy needed to remove an electron from the metal surface.

  7. Write Einstein's photoelectric equation.

    $$h\nu = h\nu_0 + \frac{1}{2}m v_{\max}^{2}$$ i.e. the photon's energy equals the work function plus the maximum kinetic energy of the emitted electron: $KE_{\max} = h(\nu - \nu_0)$.

  8. List the key experimental observations of the photoelectric effect that classical wave theory failed to explain.

    (1) Electrons are ejected only if $\nu \geq \nu_0$, regardless of intensity. (2) Kinetic energy of electrons depends on frequency, not intensity. (3) Number of electrons ejected depends on intensity. (4) Emission is instantaneous (no time lag).

  9. What type of spectrum is the hydrogen atom spectrum, and how is it produced?

    It is a line emission spectrum (discontinuous, discrete bright lines). It is produced when an electric discharge is passed through gaseous hydrogen; excited H atoms emit radiation at specific wavelengths as electrons fall to lower energy levels.

  10. Write the Rydberg formula for the hydrogen spectrum and give the Rydberg constant.

    $$\bar{\nu} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} \right)$$ with $n_2 > n_1$ and Rydberg constant $R_H = 1.097 \times 10^{7}\ \text{m}^{-1}$ (or $109677\ \text{cm}^{-1}$).

  11. Name the spectral series of hydrogen and the regions in which they appear.

    Lyman ($n_1=1$, ultraviolet), Balmer ($n_1=2$, visible), Paschen ($n_1=3$, infrared), Brackett ($n_1=4$, infrared), Pfund ($n_1=5$, far infrared).

  12. State the main postulates of Bohr's model of the hydrogen atom.

    (1) The electron revolves around the nucleus in fixed circular orbits (stationary states) without radiating energy. (2) Only orbits with angular momentum $mvr = \frac{nh}{2\pi}$ are allowed. (3) Energy is absorbed or emitted only when an electron jumps between orbits, with $\Delta E = h\nu$.

  13. Write Bohr's quantization condition for angular momentum.

    $$mvr = \frac{nh}{2\pi}, \quad n = 1, 2, 3, \dots$$ The angular momentum of the electron is quantized in integral multiples of $\frac{h}{2\pi}$.

  14. Give the expression for the radius of the $n$th Bohr orbit of hydrogen and the value for $n=1$.

    $$r_n = \frac{n^{2} h^{2} \varepsilon_0}{\pi m e^{2}} = 0.529 \times n^{2}\ \text{\AA}$$ For $n=1$ (Bohr radius), $r_1 = 0.529\ \text{\AA} = 52.9\ \text{pm}$. For hydrogen-like ions, $r_n = 0.529\,\frac{n^2}{Z}\ \text{\AA}$.

  15. Give the expression for the energy of the electron in the $n$th Bohr orbit of hydrogen.

    $$E_n = -\frac{2\pi^{2} m e^{4}}{n^{2} h^{2}} = -\frac{13.6}{n^{2}}\ \text{eV} = -\frac{2.18 \times 10^{-18}}{n^{2}}\ \text{J}$$ For hydrogen-like species, $E_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}$.

  16. Why is the energy of the electron in a Bohr orbit negative?

    The negative sign indicates that the electron is bound to (attracted by) the nucleus; its energy is lower than that of a free electron at rest at infinity, which is taken as the zero of energy. Energy must be supplied to remove the electron.

  17. How does the velocity of the electron in the $n$th Bohr orbit depend on $n$ and $Z$?

    $$v_n = \frac{2\pi e^{2}}{n h} \cdot Z \;\propto\; \frac{Z}{n}$$ For hydrogen ($Z=1$), $v_1 = 2.18 \times 10^{6}\ \text{m s}^{-1}$. Velocity decreases as $n$ increases.

  18. Define the ionization energy of hydrogen using the Bohr model.

    It is the energy required to remove the electron from $n=1$ to $n=\infty$: $$E_{\text{ion}} = E_\infty - E_1 = 0 - (-13.6\ \text{eV}) = 13.6\ \text{eV}$$ per atom ($= 1312\ \text{kJ mol}^{-1}$).

  19. List the main limitations of Bohr's model.

    (1) Fails for atoms/ions with more than one electron. (2) Cannot explain fine structure or splitting of spectral lines (Zeeman and Stark effects). (3) Violates Heisenberg's uncertainty principle by assuming fixed orbits. (4) Cannot explain chemical bonding or the wave nature of the electron.

  20. What is the dual nature of matter (wave-particle duality)?

    Matter exhibits both particle-like and wave-like properties. Just as light shows particle nature (photons) and wave nature, every moving material particle has an associated wave (matter wave). This was proposed by Louis de Broglie.

  21. Write de Broglie's relationship and define its terms.

    $$\lambda = \frac{h}{p} = \frac{h}{mv}$$ where $\lambda$ is the wavelength of the matter wave, $h$ is Planck's constant, $p = mv$ is momentum, $m$ is mass, and $v$ is velocity.

  22. Why are de Broglie wavelengths significant for electrons but negligible for macroscopic objects?

    Since $\lambda = \frac{h}{mv}$ and $h$ is extremely small, large masses give immeasurably small wavelengths. Electrons have very small mass, so their wavelengths are appreciable (comparable to atomic dimensions) and observable, e.g. in electron diffraction.

  23. How does de Broglie's relationship justify Bohr's quantization of angular momentum?

    For a stable orbit, the electron wave must form a standing wave: the orbit circumference is an integral number of wavelengths, $2\pi r = n\lambda$. Substituting $\lambda = \frac{h}{mv}$ gives $mvr = \frac{nh}{2\pi}$, exactly Bohr's quantization condition.

  24. Express the de Broglie wavelength of an electron accelerated through a potential difference $V$.

    $$\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\ \text{\AA}$$ where $V$ is in volts. This follows from equating kinetic energy $\frac{1}{2}mv^2 = eV$.

What this deck covers

The Physical Chemistry deck follows the JEE Main Physical Chemistry syllabus — 8 chapters and 53 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 6.5 cards per chapter.

Answers are written to be recallable, not just readable — averaging about 209 characters, which is long enough to carry the reasoning and short enough to say out loud.

A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.

Physical Chemistry flashcards FAQ

How many Physical Chemistry flashcards are in this JEE Main deck?

52 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.

Are these JEE Main flashcards free?

Yes. The preview here is free to read with no signup, and the full 52-card deck is free inside the Examius app.

What do the Physical Chemistry cards cover?

They follow the JEE Main Physical Chemistry syllabus — 8 chapters and 53 topics — so the questions track what is actually examinable.

How should I use these flashcards?

Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.