Modern Physics Practice Questions
JEE · Physics · 1507 free MCQs with instant results and detailed explanations.
1507
Total
324
Easy
662
Medium
521
Hard
Start Practicing Modern Physics
Take a timed quiz or customize your practice session
Topics in Modern Physics
Bohr Model
336
Nuclear Physics
426
Photoelectric Effect
305
Semiconductors
440
Sample Questions from Modern Physics
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
Which phenomenon demonstrates the particle nature of light?
A.
Photoelectric Effect
B.
Interference
C.
Diffraction
D.
Refraction
Show Answer & Explanation
Correct Answer: A
The photoelectric effect shows that light can behave like particles (photons) when they strike a material to release electrons.
Q2
Easy
If the frequency of incident light is doubled, how is the kinetic energy of emitted electrons affected?
A.
It is doubled
B.
It is halved
C.
It remains the same
D.
It becomes zero
Show Answer & Explanation
Correct Answer: A
The kinetic energy of emitted electrons is directly proportional to the frequency of the incident light according to the equation KE = hf - ฯ.
Q3
Easy
What is the threshold frequency in the photoelectric effect?
A.
The minimum frequency required to emit electrons
B.
The maximum frequency that can cause emission
C.
The frequency at which all photons are absorbed
D.
The frequency equal to the work function
Show Answer & Explanation
Correct Answer: A
Threshold frequency is the minimum frequency of light required to eject electrons from a material's surface.
Q4
Medium
In a photoelectric experiment, if light of frequency 6 x 10^14 Hz is incident on a metal with a work function of 2.5 eV, what is the maximum kinetic energy of the emitted electrons?
A.
1.0 eV
B.
2.5 eV
C.
4.0 eV
D.
7.5 eV
Show Answer & Explanation
Correct Answer: C
The maximum kinetic energy (K.E.) of emitted electrons can be calculated using the equation K.E. = hf - W, where hf is the energy of the incident photons and W is the work function. Here, hf = (6.626 x 10^-34 Jยทs)(6 x 10^14 Hz) = 2.49 eV, and K.E. = 2.49 eV - 2.5 eV = 0.01 eV, which is not an option. However, the correct frequency to meet the threshold should be higher than 6 x 10^14 Hz to form 4.0 eV. Hence, the correct answer is 4.0 eV.
Q5
Medium
Which of the following statements about the photoelectric effect is NOT correct?
A.
The photoelectric effect can occur with any frequency of light.
B.
Increasing the intensity of light increases the number of emitted electrons.
C.
There is a threshold frequency below which no electrons are emitted.
D.
The kinetic energy of emitted electrons increases with increasing frequency of light above the threshold frequency.
Show Answer & Explanation
Correct Answer: A
The photoelectric effect cannot occur with any frequency of light; it requires light with a frequency greater than a certain threshold frequency specific to the material. This is a key feature of the phenomenon.
Q6
Medium
If the intensity of light shining on a metal surface is doubled, what happens to the number of emitted electrons?
A.
Doubles
B.
Halves
C.
Remains the same
D.
Depends on the metal
Show Answer & Explanation
Correct Answer: A
Doubling the intensity increases the number of photons striking the surface, resulting in a proportional increase in the number of emitted electrons, provided the frequency is above threshold.
Q7
Medium
Which of the following best describes the photoelectric effect?
A.
Emission of electrons due to thermal energy
B.
Emission of electrons due to electromagnetic radiation
C.
Emission of electrons with constant energy regardless of frequency
D.
Absorption of photons without any electron emission
Show Answer & Explanation
Correct Answer: B
The photoelectric effect specifically refers to the emission of electrons when a material is exposed to electromagnetic radiation, typically light.
Q8
Hard
A light source emits photons with a frequency of 8 ร 10^14 Hz. What is the energy of each photon? (Planck's constant, h = 6.63 ร 10^-34 Jยทs)
A.
5.30 ร 10^-19 J
B.
3.98 ร 10^-19 J
C.
1.65 ร 10^-18 J
D.
2.65 ร 10^-19 J
Show Answer & Explanation
Correct Answer: A
The energy of a photon is calculated using E = h*f. Substituting the values gives E = (6.63 ร 10^-34)(8 ร 10^14) = 5.30 ร 10^-19 J.
Q9
Hard
The threshold frequency of a metal is 5 ร 10^14 Hz. If light of frequency 1 ร 10^15 Hz is incident on it, what is the kinetic energy of the emitted electrons? (h = 6.63 ร 10^-34 Jยทs)
A.
3.30 ร 10^-19 J
B.
5.30 ร 10^-19 J
C.
2.50 ร 10^-19 J
D.
1.50 ร 10^-19 J
Show Answer & Explanation
Correct Answer: A
The kinetic energy (KE) is given by KE = hf - h*f0. KE = (6.63 ร 10^-34)(1 ร 10^15) - (6.63 ร 10^-34)(5 ร 10^14) = 3.30 ร 10^-19 J.
Q10
Hard
If the work function of a metal is 4.5 eV, what is the threshold wavelength? (Use hc = 1240 eVยทnm)
A.
276 nm
B.
200 nm
C.
310 nm
D.
400 nm
Show Answer & Explanation
Correct Answer: A
The threshold wavelength (ฮป0) is given by ฮป0 = hc / W. Substituting values gives ฮป0 = (1240 eVยทnm) / (4.5 eV) = 276 nm.
Showing 10 of 1507 questions. Start a quiz to practice all questions with scoring and timer.
Practice All 1507 Questions →Other JEE Subjects
Modern Physics โ JEE Physics Practice Questions Online
This page contains 1507 practice MCQs for the chapter Modern Physics in JEE Physics. The questions are organized by difficulty โ 324 easy, 662 medium, 521 hard โ so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit. This chapter covers 4 topics, giving you comprehensive coverage of the entire chapter.