MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4d Light Electromagnetic Radiation

What this deck covers

This deck focuses on 4d Light Electromagnetic Radiation, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify the wavelength of a $5.0\times 10^{14}\ \text{Hz}$ photon in vacuum using $c=3.0\times 10^8\ \text{m/s}$.

Answer: $\lambda = 6.0\times 10^{-7}\ \text{m}$. Wavelength is calculated as speed of light divided by frequency, yielding the given value.

Flashcard 2: When light enters a medium with higher refractive index, what changes: speed, frequency, wavelength?

Answer: Speed decreases; frequency stays constant; wavelength decreases. Upon entering a denser medium, light speed decreases while frequency remains constant, resulting in shorter wavelength.

Flashcard 3: What is the relationship between wavelength, frequency, and wave speed for light in vacuum?

Answer: $c = \lambda f$. The wave speed in vacuum equals the product of wavelength and frequency for electromagnetic radiation.

Flashcard 4: What type of wave is electromagnetic radiation in terms of field orientation and direction of travel?

Answer: A transverse wave; $\vec{E}$ and $\vec{B}$ are perpendicular to propagation. Electromagnetic waves are transverse because the electric and magnetic fields oscillate perpendicular to the direction of wave propagation.

Flashcard 5: Light goes from air $n_1=1.00$ into glass $n_2=1.50$. What is the speed in glass using $c=3.0\times 10^8\ \text{m/s}$?

Answer: $v = 2.0\times 10^8\ \text{m/s}$. Speed in the medium is vacuum speed divided by the refractive index of glass, yielding the given velocity.

Flashcard 6: A photon has energy $4.0\ \text{eV}$. What is its wavelength in nm using $hc=1240\ \text{eV}\cdot\text{nm}$?

Answer: $\lambda = 310\ \text{nm}$. Wavelength in nm is found by dividing $hc$ by energy in eV, producing the indicated wavelength.

Flashcard 7: What is the photon energy in eV for $\lambda = 620\ \text{nm}$ using $hc=1240\ \text{eV}\cdot\text{nm}$?

Answer: $E = 2.0\ \text{eV}$. Photon energy in eV is obtained by dividing $hc$ by wavelength in nm, giving the stated energy.

Flashcard 8: Identify the frequency of $600\ \text{nm}$ light in vacuum using $c=3.0\times 10^8\ \text{m/s}$.

Answer: $f = 5.0\times 10^{14}\ \text{Hz}$. Frequency is determined by dividing speed of light by wavelength, resulting in the specified value.

Flashcard 9: State Einstein's photoelectric equation for maximum kinetic energy of emitted electrons.

Answer: $K_{\max} = hf - \phi$. In the photoelectric effect, maximum kinetic energy is incident photon energy minus the work function.

Flashcard 10: What is the work function relationship for photoelectric emission in terms of threshold frequency?

Answer: $\phi = hf_0$. The work function equals the energy of photons at the threshold frequency for electron emission.

Flashcard 11: What is the photon momentum in terms of energy for light in vacuum?

Answer: $p = \frac{E}{c}$. For photons, relativistic energy-momentum relation simplifies to momentum as energy divided by speed of light.

Flashcard 12: Identify the formula for the momentum of a photon in terms of wavelength.

Answer: $p = \frac{h}{\lambda}$. Photon momentum arises from wave-particle duality, equaling Planck's constant divided by wavelength.

Flashcard 13: What is the intensity trend with distance for a point light source in free space?

Answer: $I \propto \frac{1}{r^2}$. Intensity follows the inverse square law as power spreads over the surface area of an expanding sphere.

Flashcard 14: What is the relationship between frequency and wavelength for light in a fixed medium?

Answer: $f \propto \frac{1}{\lambda}$ because $v = \lambda f$ with fixed $v$. With constant wave speed in a medium, frequency and wavelength maintain an inverse proportionality.

Flashcard 15: State the critical angle formula for total internal reflection from $n_1$ to $n_2$ with $n_1>n_2$.

Answer: $\sin\theta_c = \frac{n_2}{n_1}$. The critical angle occurs when the refracted angle is 90 degrees, leading to total internal reflection for larger incident angles.

Flashcard 16: What is the condition for total internal reflection in terms of indices of refraction?

Answer: Light must go from higher to lower $n$: $n_1 > n_2$. Total internal reflection requires light traveling from a medium of higher refractive index to one of lower index.

Flashcard 17: What is the approximate speed of light in vacuum used for MCAT calculations?

Answer: $c \approx 3.0 \times 10^8\ \text{m/s}$. This value represents the constant speed at which light travels in vacuum, fundamental for wave calculations.

Flashcard 18: State Snell's law relating incident and refracted angles and refractive indices.

Answer: $n_1\sin\theta_1 = n_2\sin\theta_2$. Snell's law governs refraction by equating the products of refractive indices and sines of angles across an interface.

Flashcard 19: What is the index of refraction in terms of light speeds in vacuum and in a medium?

Answer: $n = \frac{c}{v}$. Refractive index quantifies the reduction in light speed within a medium relative to vacuum.

Flashcard 20: What happens to photon energy when wavelength decreases?

Answer: Photon energy increases because $E \propto \frac{1}{\lambda}$. Photon energy is inversely proportional to wavelength, so decreasing wavelength increases energy.

Flashcard 21: Which option correctly orders electromagnetic radiation from lowest to highest frequency?

Answer: Radio < microwave < IR < visible < UV < X-ray < gamma. The electromagnetic spectrum is ordered by increasing frequency, which corresponds to increasing energy and decreasing wavelength.

Flashcard 22: What is $hc$ in convenient units for spectroscopy calculations?

Answer: $hc \approx 1240\ \text{eV}\cdot\text{nm}$. This product enables straightforward computation of photon energy in electronvolts from wavelength in nanometers.

Flashcard 23: What is Planck's constant to two significant figures for MCAT use?

Answer: $h \approx 6.6 \times 10^{-34}\ \text{J}\cdot\text{s}$. This approximate value of Planck's constant is used in quantum mechanics calculations for energy-frequency relations.

Flashcard 24: State the formula for photon energy in terms of wavelength.

Answer: $E = \frac{hc}{\lambda}$. Derived from $E=hf$ and $c=lambda f$, photon energy is inversely proportional to wavelength.

Flashcard 25: State the formula for photon energy in terms of frequency.

Answer: $E = hf$. Photon energy is directly proportional to its frequency, with Planck's constant as the proportionality factor.