What is Modern physics on the Physics Regents?

Modern physics is a Modern Physics topic on the Physics Regents. RegentsLab has 42 practice questions on it, each with a full worked solution.

How do you study Modern physics for the Physics Regents?

Work a range of Modern physics question types and read the worked solution every time you miss one — the exam reuses the same ideas year after year. Start with the questions on this page.

Modern physics: Physics Regents Practice

Practice Modern physics for the Physics Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

Open the practice lab Past exams

Question 81 · 2 pts

A photon traveling through a vacuum has a frequency of $5.0 \times 10^{14}$ Hz. What is the energy of this photon?

(A) $1.3 \times 10^{-48}$ J
(B) $7.5 \times 10^{47}$ J
(C) $3.3 \times 10^{-19}$ J ✓
(D) $3.3 \times 10^{-20}$ J

Worked solution

$E = hf$ Photon energy equation from the Modern Physics section of the reference tables.

$E = (6.63 \times 10^{-34})(5.0 \times 10^{14})$

$E = 3.315 \times 10^{-19} \approx 3.3 \times 10^{-19}\ \text{J}$

Answer: $3.3 \times 10^{-19}$ J

A photon's energy is directly proportional to its frequency, $E = hf$, using Planck's constant $h = 6.63 \times 10^{-34}$ J·s from the reference tables.

Question 87 · 2 pts

According to the Standard Model of particle physics, a proton is composed of which combination of quarks?

(A) two up quarks and one strange quark
(B) two up quarks and one down quark ✓
(C) one up quark and two down quarks
(D) three up quarks

Worked solution

$p = u + u + d$ A proton is a baryon, built from three quarks.

$q = +\tfrac{2}{3} + \tfrac{2}{3} - \tfrac{1}{3} = +1$ The quark charges sum to the proton's +1 charge, confirming uud.

Answer: two up quarks and one down quark

In the Standard Model, a proton is a baryon made of two up quarks and one down quark ($uud$), whose charges $+\tfrac{2}{3}, +\tfrac{2}{3}, -\tfrac{1}{3}$ add to a net charge of $+1$.

Question 96 · 2 pts

According to the Standard Model classification of matter, a particle composed of one quark and one antiquark is classified as a

(A) lepton
(B) baryon
(C) photon
(D) meson ✓

Worked solution

$\text{hadrons} \rightarrow \text{baryons (3 quarks), mesons (quark + antiquark)}$ From the classification-of-matter chart.

$\text{quark} + \text{antiquark} \Rightarrow \text{meson}$

Answer: meson

Hadrons are quark composites: baryons contain three quarks, while mesons consist of a quark paired with an antiquark.

Question 64 · 4 pts · constructed response

A photon has a frequency of $5.0 \times 10^{14}\ \text{Hz}$. Calculate the energy of the photon in joules. (Planck's constant $h = 6.63 \times 10^{-34}\ \text{J·s}$.)

Model answer: The photon's energy is approximately 3.3 x 10^-19 J.

Worked solution

$E = hf$ Photon energy equation

$E = (6.63 \times 10^{-34})(5.0 \times 10^{14})$

$E = 3.315 \times 10^{-19}\ \text{J}$

$E \approx 3.3 \times 10^{-19}\ \text{J}$ Round to two significant figures

Answer: The photon's energy is approximately 3.3 x 10^-19 J.

The energy of a photon is directly proportional to its frequency through Planck's constant, so multiplying the given frequency by h yields the photon's energy.

Question 37 · 2 pts

A photon has a frequency of $5.0\times10^{14}\ \text{Hz}$. Using Planck's constant $h = 6.63\times10^{-34}\ \text{J}\cdot\text{s}$, what is the energy of this photon?

(A) $3.3\times10^{-19}$ J ✓
(B) $6.6\times10^{-19}$ J
(C) $1.3\times10^{-48}$ J
(D) $1.3\times10^{-19}$ J

Worked solution

$E = hf$

$E = (6.63\times10^{-34}\ \text{J}\cdot\text{s})(5.0\times10^{14}\ \text{Hz})$

$E = 3.3\times10^{-19}\ \text{J}$

Answer: $3.3\times10^{-19}$ J

Photon energy is given by $E=hf$; multiplying Planck's constant by the frequency gives $3.3\times10^{-19}$ J.

Question 82 · 2 pts

Light of wavelength $4.5 \times 10^{-7}$ m travels through a vacuum. What is the energy of one photon of this light?

(A) $4.4 \times 10^{-19}$ J ✓
(B) $3.0 \times 10^{-40}$ J
(C) $9.0 \times 10^{-32}$ J
(D) $6.7 \times 10^{14}$ J

Worked solution

$E = \frac{hc}{\lambda}$ Combine E = hf with f = c/lambda.

$E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{4.5 \times 10^{-7}}$

$E = \frac{1.989 \times 10^{-25}}{4.5 \times 10^{-7}} = 4.4 \times 10^{-19}\ \text{J}$

Answer: $4.4 \times 10^{-19}$ J

Because $f = c/\lambda$, photon energy is $E = hc/\lambda$; shorter wavelengths of light carry more energy per photon.

Question 83 · 2 pts

In a hydrogen atom, an electron drops from the $n=3$ energy level ($-1.51$ eV) to the $n=2$ energy level ($-3.40$ eV). What is the energy of the emitted photon?

(A) $1.51$ eV
(B) $4.91$ eV
(C) $-1.89$ eV
(D) $1.89$ eV ✓

Worked solution

$E_{photon} = E_i - E_f$ Emitted photon energy equals the difference between the initial and final levels.

$E_{photon} = (-1.51\ \text{eV}) - (-3.40\ \text{eV})$

$E_{photon} = 1.89\ \text{eV}$

Answer: $1.89$ eV

When an electron falls between energy levels, the emitted photon carries energy equal to the positive difference $E_i - E_f$ shown on the hydrogen energy-level diagram.

Question 86 · 2 pts

A hydrogen atom in its ground state ($-13.60$ eV) absorbs a single photon and becomes ionized. What is the minimum energy, in joules, that the absorbed photon must have?

(A) $2.18 \times 10^{-18}$ J ✓
(B) $5.44 \times 10^{-19}$ J
(C) $2.18 \times 10^{-19}$ J
(D) $8.50 \times 10^{19}$ J

Worked solution

$\Delta E = 0.00\ \text{eV} - (-13.60\ \text{eV}) = 13.60\ \text{eV}$ Energy needed to raise the electron from the ground state to ionization (0.00 eV).

$E = (13.60\ \text{eV})(1.60 \times 10^{-19}\ \text{J/eV})$

$E = 2.18 \times 10^{-18}\ \text{J}$

Answer: $2.18 \times 10^{-18}$ J

Ionizing ground-state hydrogen requires a photon whose energy equals the $13.60$ eV binding energy, converted to joules with $1\ \text{eV} = 1.60 \times 10^{-19}$ J.

Question 88 · 2 pts

Compared to a photon of red light, a photon of violet light has

(A) less energy and a lower frequency
(B) greater energy and a higher frequency ✓
(C) less energy and a higher frequency
(D) greater energy and a lower frequency

Worked solution

$E = hf$ Photon energy is directly proportional to frequency.

$f_{violet} > f_{red} \Rightarrow E_{violet} > E_{red}$ Violet light has a higher frequency and shorter wavelength than red light.

Answer: greater energy and a higher frequency

By $E = hf$, a higher frequency means greater photon energy; violet light has both a higher frequency and a shorter wavelength than red light.

Question 89 · 2 pts

A photon of light has a frequency of $5.0 \times 10^{14}\ \mathrm{Hz}$. Using the reference-table value $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$, what is the energy of the photon?

(A) $1.3 \times 10^{-48}\ \mathrm{J}$
(B) $3.3 \times 10^{-20}\ \mathrm{J}$
(C) $1.5 \times 10^{23}\ \mathrm{J}$
(D) $3.3 \times 10^{-19}\ \mathrm{J}$ ✓

Worked solution

$E_{photon} = hf$ Photon energy relation from the reference tables.

$E = (6.63 \times 10^{-34})(5.0 \times 10^{14})$

$E = 33.15 \times 10^{-20} = 3.3 \times 10^{-19}\ \mathrm{J}$

Answer: $3.3 \times 10^{-19}\ \mathrm{J}$

A photon's energy is directly proportional to its frequency through Planck's constant, $E = hf$.

Question 90 · 2 pts

Ultraviolet light has a wavelength of $3.0 \times 10^{-7}\ \mathrm{m}$. Using $h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s}$ and $c = 3.0 \times 10^{8}\ \mathrm{m/s}$, what is the energy of one of its photons?

(A) $1.0 \times 10^{15}\ \mathrm{J}$
(B) $2.2 \times 10^{-27}\ \mathrm{J}$
(C) $6.0 \times 10^{-32}\ \mathrm{J}$
(D) $6.6 \times 10^{-19}\ \mathrm{J}$ ✓

Worked solution

$E_{photon} = \frac{hc}{\lambda}$ Combine $E = hf$ with $c = f\lambda$.

$E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{3.0 \times 10^{-7}}$

$E = \frac{1.989 \times 10^{-25}}{3.0 \times 10^{-7}} = 6.6 \times 10^{-19}\ \mathrm{J}$

Answer: $6.6 \times 10^{-19}\ \mathrm{J}$

Shorter wavelengths carry more energy per photon because $E = hc/\lambda$ is inversely proportional to wavelength.

Question 92 · 2 pts

In a hydrogen atom, an electron drops from the $n=3$ energy level ($-1.51\ \mathrm{eV}$) to the $n=2$ level ($-3.40\ \mathrm{eV}$). Using $E_{photon} = E_i - E_f$ from the reference tables, what is the energy of the emitted photon?

(A) $1.89\ \mathrm{eV}$ ✓
(B) $10.20\ \mathrm{eV}$
(C) $4.91\ \mathrm{eV}$
(D) $12.09\ \mathrm{eV}$

Worked solution

$E_{photon} = E_i - E_f$ Emission: energy lost by the electron becomes the photon.

$E = (-1.51) - (-3.40)$

$E = -1.51 + 3.40 = 1.89\ \mathrm{eV}$

Answer: $1.89\ \mathrm{eV}$

When an electron falls to a lower energy level, it emits a photon whose energy equals the difference between the two levels.

Question 95 · 2 pts

According to the Standard Model, the up quark has a charge of $+\tfrac{2}{3}e$ and the down quark has a charge of $-\tfrac{1}{3}e$. A particle is made of one up quark and two down quarks. What is its net charge, in elementary charges $e$?

(A) $+1$
(B) $+\tfrac{4}{3}$
(C) $-1$
(D) $0$ ✓

Worked solution

$q = q_{up} + q_{down} + q_{down}$ Sum the quark charges from the Standard Model chart.

$q = \left(+\tfrac{2}{3}\right) + \left(-\tfrac{1}{3}\right) + \left(-\tfrac{1}{3}\right)$

$q = \tfrac{2}{3} - \tfrac{2}{3} = 0$

Answer: $0$

A particle built from one up and two down quarks has zero net charge; this combination is a neutron.

Question 11 · 2 pts

A photon has a frequency of $6.0\times10^{14}\ \text{Hz}$. Using Planck's constant $h = 6.63\times10^{-34}\ \text{J}\cdot\text{s}$, what is the energy of the photon?

(A) $4.0\times10^{-19}\ \text{J}$ ✓
(B) $1.1\times10^{-48}\ \text{J}$
(C) $4.0\times10^{-48}\ \text{J}$
(D) $3.6\times10^{-19}\ \text{J}$

Worked solution

$E = hf$ Photon energy equation

$E = (6.63\times10^{-34}\ \text{J}\cdot\text{s})(6.0\times10^{14}\ \text{Hz})$

$E = 39.78\times10^{-20}\ \text{J} = 4.0\times10^{-19}\ \text{J}$

Answer: $4.0\times10^{-19}\ \text{J}$

Photon energy equals Planck's constant times frequency; multiplying the mantissas and adding the exponents (-34 + 14 = -20) and adjusting gives 4.0 x 10^-19 J.

Question 24 · 2 pts

What is the energy of a photon of light that has a frequency of $5.0\times10^{14}\text{ Hz}$? (Planck's constant $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$)

(A) $1.3\times10^{-19}\text{ J}$
(B) $6.6\times10^{-34}\text{ J}$
(C) $3.3\times10^{-13}\text{ J}$
(D) $3.3\times10^{-19}\text{ J}$ ✓

Worked solution

$E = hf$ Photon energy equation

$E = (6.63\times10^{-34}\text{ J}\cdot\text{s})(5.0\times10^{14}\text{ Hz})$

$E = 3.3\times10^{-19}\text{ J}$

Answer: $3.3\times10^{-19}\text{ J}$

The energy of a photon is found by multiplying Planck's constant by the light's frequency, giving 3.3x10^-19 J.

Question 38 · 2 pts

What is the energy of a photon of light having a frequency of $5.0 \times 10^{14}\ \text{Hz}$? (Planck's constant $h = 6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$)

(A) 3.3 × 10⁻¹⁹ J ✓
(B) 3.3 × 10⁻²⁰ J
(C) 1.3 × 10⁻⁴⁸ J
(D) 6.6 × 10⁻³⁴ J

Worked solution

$E = hf$ Photon energy equation

$E = (6.63\times10^{-34}\ \text{J}\cdot\text{s})(5.0\times10^{14}\ \text{Hz})$

$E = 33.15\times10^{-20}\ \text{J} = 3.3\times10^{-19}\ \text{J}$

Answer: 3.3 × 10⁻¹⁹ J

The energy of a photon is given by E = hf, the product of Planck's constant and the light's frequency. Multiplying the coefficients and adding the exponents (−34 + 14 = −20) gives 33.15 × 10⁻²⁰ J, which correctly rewrites to 3.3 × 10⁻¹⁹ J; failing to renormalize this power of ten gives the 3.3 × 10⁻²⁰ J distractor.

Question 50 · 2 pts

A photon of electromagnetic radiation has a frequency of 6.0 x 10^14 Hz. What is the energy of this photon? (Planck's constant h = 6.63 x 10^-34 J·s.)

(A) 6.63 x 10^-34 J
(B) 1.1 x 10^-48 J
(C) 4.0 x 10^-19 J ✓
(D) 3.98 x 10^-20 J

Worked solution

$E = hf$

$E = (6.63\times10^{-34}\ \text{J·s})(6.0\times10^{14}\ \text{Hz})$

$E = 39.78\times10^{-20}\ \text{J} = 3.978\times10^{-19}\ \text{J}$

$E \approx 4.0\times10^{-19}\ \text{J}$

Answer: 4.0 x 10^-19 J

Photon energy is found from the Planck relation E = hf, multiplying the given frequency by Planck's constant and combining the powers of ten correctly to get about 4.0 x 10^-19 J. Mishandling the powers of ten (leaving the answer as 39.78 x 10^-20 without renormalizing) or dividing instead of multiplying produces the distractor values.

Question 12 · 2 pts

What is the energy of a photon whose frequency is $5.0 \times 10^{14}$ Hz? (Planck's constant $h = 6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$)

(A) $6.63 \times 10^{-34}$ J
(B) $3.3 \times 10^{-19}$ J ✓
(C) $1.3 \times 10^{-48}$ J
(D) $1.3 \times 10^{15}$ J

Worked solution

$E = hf$ Planck's equation for photon energy

$E = (6.63\times10^{-34}\text{ J·s})(5.0\times10^{14}\text{ Hz})$

$E \approx 3.3\times10^{-19}\text{ J}$

Answer: $3.3 \times 10^{-19}$ J

A photon's energy equals Planck's constant times its frequency, E = hf. Multiplying 6.63x10^-34 J.s by 5.0x10^14 Hz gives about 3.3x10^-19 J; adding exponents incorrectly, using only h, or dividing instead of multiplying produces the distractor values.

Question 24 · 2 pts

What is the energy of a photon that has a frequency of $5.0\times10^{14}\ \text{Hz}$? (Use $h = 6.63\times10^{-34}\ \text{J}\cdot\text{s}$.)

(A) $1.33\times10^{-48}$ J
(B) $6.63\times10^{-34}$ J
(C) $3.3\times10^{-19}$ J ✓
(D) $3.3\times10^{-20}$ J

Worked solution

$E = hf$

$E = (6.63\times10^{-34}\ \text{J}\cdot\text{s})(5.0\times10^{14}\ \text{Hz})$

$E = 3.315\times10^{-19}\ \text{J} \approx 3.3\times10^{-19}\ \text{J}$

Answer: $3.3\times10^{-19}$ J

The energy of a photon is directly proportional to its frequency through Planck's constant, $E=hf$. A common error is a decimal placement mistake when multiplying the powers of ten, or dividing $h$ by $f$ instead of multiplying.

Question 36 · 2 pts

A photon has a frequency of $6.0 \times 10^{14}\ \text{Hz}$. What is the energy of this photon? (Planck's constant $h = 6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$)

(A) $9.05 \times 10^{47}\ \text{J}$
(B) $1.11 \times 10^{-48}\ \text{J}$
(C) $3.98 \times 10^{-19}\ \text{J}$ ✓
(D) $3.98 \times 10^{-20}\ \text{J}$

Worked solution

$E = hf$ Photon energy equation

$E = (6.63 \times 10^{-34}\ \text{J}\cdot\text{s})(6.0 \times 10^{14}\ \text{Hz})$

$E = 3.98 \times 10^{-19}\ \text{J}$

Answer: $3.98 \times 10^{-19}\ \text{J}$

A photon's energy is directly proportional to its frequency, given by $E=hf$; a common error is dividing instead of multiplying, which gives an unreasonably small or large result.

Question 48 · 2 pts

What is the energy of a photon of light with a frequency of 6.0 x 10^14 Hz? (Planck's constant, h = 6.63 x 10^-34 J*s)

(A) 4.0 x 10^-19 J ✓
(B) 1.1 x 10^-48 J
(C) 3.98 x 10^-15 J
(D) 6.6 x 10^-34 J

Worked solution

$E = hf$

$E = (6.63\times10^{-34}\text{ J·s})(6.0\times10^{14}\text{ Hz})$

$E = 3.98\times10^{-19}\text{ J} \approx 4.0\times10^{-19}\text{ J}$

Answer: 4.0 x 10^-19 J

The energy of a photon is given by Planck's equation, E = hf; multiplying Planck's constant by the given frequency yields a photon energy of approximately 4.0 x 10^-19 J, consistent with visible-light photon energies.

Question 58 · 4 pts · constructed response

Light with a frequency of $5.00\times10^{14}\ \text{Hz}$ strikes a metal surface. Using Planck's constant $h = 6.63\times10^{-34}\ \text{J}\cdot\text{s}$, calculate the energy of a single photon of this light in joules.

Model answer: The energy of one photon is about 3.32 x 10^-19 J.

Worked solution

$E = hf$ Planck's equation for photon energy

$E = (6.63\times10^{-34})(5.00\times10^{14})$

$E = 3.315\times10^{-19}\ \text{J} \approx 3.32\times10^{-19}\ \text{J}$

Answer: The energy of one photon is about 3.32 x 10^-19 J.

Photon energy depends only on its frequency through Planck's constant, reflecting the quantized nature of electromagnetic radiation described by modern physics.

Question 27 · 2 pts

A particle with rest mass $1.0 \times 10^{-27}\ \text{kg}$ is completely converted into energy. How much energy is released? (speed of light $c = 3.00 \times 10^8\ \text{m/s}$)

(A) $9.0 \times 10^{-11}\ \text{J}$ ✓
(B) $3.0 \times 10^{8}\ \text{J}$
(C) $3.0 \times 10^{-19}\ \text{J}$
(D) $9.0 \times 10^{-19}\ \text{J}$

Worked solution

$E = mc^2$ mass-energy equivalence

$E = (1.0 \times 10^{-27}\ \text{kg})(3.00 \times 10^{8}\ \text{m/s})^2$

$E = (1.0 \times 10^{-27}\ \text{kg})(9.00 \times 10^{16}\ \text{m}^2/\text{s}^2)$

$E = 9.0 \times 10^{-11}\ \text{J}$

Answer: $9.0 \times 10^{-11}\ \text{J}$

Einstein's mass-energy equivalence states that energy equals mass times the speed of light squared; squaring c before multiplying by the mass is essential to obtain the correct energy released.

Question 26 · 2 pts

A photon of electromagnetic radiation has a frequency of $6.0 \times 10^{14}\ \text{Hz}$. What is the energy of this photon?

(A) $9.11 \times 10^{-31}\ \text{J}$
(B) $1.1 \times 10^{-48}\ \text{J}$
(C) $1.66 \times 10^{19}\ \text{J}$
(D) $3.98 \times 10^{-19}\ \text{J}$ ✓

Worked solution

$E = hf$ Planck's equation, $h = 6.63\times10^{-34}\ \text{J·s}$

$E = (6.63\times10^{-34}\ \text{J\cdot s})(6.0\times10^{14}\ \text{Hz}) = 3.98\times10^{-19}\ \text{J}$

Answer: $3.98 \times 10^{-19}\ \text{J}$

Photon energy is directly proportional to frequency through Planck's constant, $E = hf$. Multiplying Planck's constant by the given frequency yields the photon's energy in joules.

Showing 24 of 42 Modern physics questions — practice the rest free in the practice lab.

Modern physics: Physics Regents Practice

More Physics topics

More Physics help