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NEET 45-Question Mock 45:00

Q1. In a photoelectric experiment, the intensity of incident radiation is doubled while the frequency is kept constant (above threshold). What will happen?

Explanation:
Option C is correct. When intensity increases, the number of incident photons per second increases. This leads to a greater number of electrons being emitted from the metal surface. Higher intensity → more photons → more emitted electrons → higher current

Q2. An electron is accelerated through a potential difference of 100 V. Its de Broglie wavelength is:

Explanation:
Option A is correct. Use the standard shortcut formula for electrons: λ(A˚)= 12.27/_/¯v , For V=100 , V: root(100) = 10 , λ=12.27/10, ​=1.227, A˚≈1.23 A˚

Q3. Which of the following observations of the photoelectric effect cannot be explained by the wave theory of light?

Explanation:
Option B is correct. The wave theory of light fails to explain the existence of a threshold frequency because, According to wave theory, energy depends on intensity, not frequency. So, electrons should be emitted at any frequency if intensity is high enough. But experimentally, no electrons are emitted below a minimum frequency, no matter how high the intensity is.

Q4. If the wavelength of incident light decreases while intensity remains constant, what happens to emitted electrons?

Explanation:
Option B is correct. Energy of a photon is given by: E=hc/λ So, when wavelength λ decreases, photon energy increases. Decreasing wavelength ⇒ higher photon energy Work function ϕ remains constant (property of metal) Therefore, maximum kinetic energy of emitted electrons increases

Q5. Two metals A and B have work functions 2 eV and 4 eV respectively. Which statement is correct?

Explanation:
Option A is correct. The threshold frequency (ν₀) is the minimum frequency of incident radiation required to eject electrons from a metal surface. It is given by the relation: ν₀ = φ / h , where φ is the work function of the metal and h is Planck’s constant. , Since ν₀ is directly proportional to the work function (φ), a metal with a higher work function will have a higher threshold frequency. , Given: Metal A: φ = 2 eV , Metal B: φ = 4 eV , Metal B has a higher work function than Metal A, so Metal B also has a higher threshold frequency.

Q6. The de Broglie wavelength of a particle is inversely proportional to:

Explanation:
Option C is correct. The de Broglie relation is: λ=h/p where: λ = de Broglie wavelength, h = Planck’s constant, p = momentum, So, λ∝1/p Therefore, the de Broglie wavelength is inversely proportional to momentum.​

Q7. In the photoelectric effect, emission of electrons takes place only when:

Explanation:
Option D is correct. For photoelectric emission to occur, the incident photon must have energy at least equal to the work function of the metal: hν≥ϕ , This implies: ν≥ν0​ , where: ν0 = threshold frequency, ϕ = work function. So, electrons are emitted only when the frequency of incident light exceeds the threshold frequency, regardless of intensity.

Q8. A photon and an electron have the same energy. Which one will have greater momentum?

Explanation:
Option B is correct. Photon: p=E/c , Electron: p=(root)2mE , Since electron has mass → momentum is greater.

Q9. Assertion (A): Photoelectric emission is instantaneous. Reason (R): Energy transfer from light is continuous.

Explanation:
Option C is correct. Assertion true → Photoelectric emission is essentially instantaneous because electrons are emitted immediately when light of sufficient frequency falls on a metal surface. There is no measurable time lag. Reason false → Energy transfer from light is not continuous. According to Einstein’s quantum theory, light energy is transferred in discrete packets called photons, not continuously.

Q10. If the velocity of an electron increases, its de Broglie wavelength:

Explanation:
Option B is correct. According to the de Broglie relation: λ=h/mv, where:λ = wavelength, h = Planck’s constant, m = mass of the electron, v = velocity, since λ∝1/v , when velocity increases, the wavelength decreases.

Q11. Stopping potential depends on:

Explanation:
Option A is correct. Stopping potential is directly related to the maximum kinetic energy of the emitted photoelectrons. According to the photoelectric equation: Kmax​=hν−ϕ, where: hν = energy of the incident photon (depends on frequency ν), ϕ = work function of the metal, Since stopping potential is proportional to Kmax, it depends only on the frequency of light, not on intensity, distance, or area.

Q12. Energy of a photon is directly proportional to:

Explanation:
Option B is correct. The energy of a photon is given by: E=hν, where: E = energy of the photon, h = Planck’s constant, ν = frequency, Thus, energy is directly proportional to frequency and inversely proportional to wavelength (E=hc/λ).

Q13. The work function of a metal depends on:

Explanation:
Option C is correct. The work function (ϕ) is the minimum energy required to remove an electron from the surface of a metal. It is an intrinsic property of the material and therefore depends on the nature of the metal. It does not depend on the intensity or frequency of incident light (though these affect emission), and temperature has only a negligible effect under normal conditions.

Q14. The radius of the first Bohr orbit of hydrogen atom is 0.53 Å. What will be the radius of the second orbit?

Explanation:
Option C is correct. According to the Bohr model, the radius of the nth orbit is given by: “The radius of the nth orbit is equal to n squared times the radius of the first orbit.” , rₙ = n² × r₁, where r1=0.53A˚ , For the second orbit (n=2): r2​=2²×0.53, =4×0.53=2.12A˚

Q15. The energy required to excite an electron in a hydrogen atom from the ground state (n=1) to n=2 is:

Explanation:
Option A is correct. The energy of an electron in the nth orbit of a hydrogen atom is given by: En​=−13.6​eV/n², So, Energy required for excitation: ΔE=E2​−E1​=(−3.4)−(−13.6)=10.2eV , Thus, the required energy is 10.2 eV.

Q16. How many spectral lines are produced when an electron in a hydrogen atom falls from n=4 to the ground state?

Explanation:
Option D is correct. The number of spectral lines produced when an electron transitions from level n to the ground state is given by: Number of lines = n(n − 1) / 2, For n = 4: , Number of lines = 4(4 − 1) / 2 , = 4 × 3 / 2 , = 12 / 2 , = 6

Q17. Which of the following series of hydrogen spectrum lies in the visible region?

Explanation:
Option B is correct. Different spectral series of the hydrogen atom correspond to transitions ending at different energy levels: Lyman series → Ultraviolet (UV) region (transitions to n=1) , Balmer series → Visible region (transitions to n=2) , Paschen series → Infrared (IR) region (transitions to n=3) , Brackett series → Infrared (IR) region (transitions to n=4) ,

Q18. According to Bohr’s model, the angular momentum of an electron is:

Explanation:
Option B is correct. According to Bohr’s quantization condition, the angular momentum of an electron in an orbit is given by: L = nh / 2π , where: L = angular momentum , n = principal quantum number , h = Planck’s constant , Thus, angular momentum is quantized and equals nh/2π.

Q19. The energy of an electron in hydrogen atom increases with:

Explanation:
Option A is correct. The energy of an electron in a hydrogen atom is given by: Eₙ = −13.6 / n² , As the principal quantum number n increases, the energy becomes less negative (i.e., increases and approaches zero). Hence, the energy of the electron increases with increase in n.

Q20. The ground state of a hydrogen atom corresponds to:

Explanation:
Option B is correct. The ground state is the lowest energy state of the hydrogen atom. In the Bohr model, the lowest allowed principal quantum number is: n = 1

Q21. The ionization energy of hydrogen atom is:

Explanation:
Option C is correct. Ionization energy is the energy required to remove an electron completely from the ground state (n = 1) of a hydrogen atom to infinity (n = ∞). , For hydrogen: Ground state energy = −13.6 eV , At infinity, energy = 0 eV , So, required energy: E=0−(−13.6)=13.6 eV

Q22. The frequency of radiation emitted during an electronic transition depends on:

Explanation:
Option C is correct. When an electron moves from a higher energy level (E₂) to a lower energy level (E₁), it emits a photon. The energy of the emitted photon is equal to the difference between the two energy levels, given by the relation: hν = E₂ − E₁ , the frequency (ν) of the emitted radiation depends only on the energy difference between the initial and final states. It does not depend on the individual values of energy levels or the path followed by the electron during the transition.

Q23. Energy of an electron in a hydrogen atom is proportional to:

Explanation:
Option C is correct. The energy of an electron in a hydrogen atom in the nth orbit is given by the formula: En = −13.6 / n² eV , it is clear that the energy is inversely proportional to the square of the principal quantum number (n). As the value of n increases, the magnitude of energy decreases and approaches zero. Therefore, the energy of the electron is proportional to 1/n².

Q24. Mass defect in a nucleus is due to:

Explanation:
Option B is correct. Mass defect is the difference between the mass of a nucleus and the sum of the masses of its individual nucleons. This missing mass is converted into nuclear binding energy according to Einstein’s equation: E = Δmc² , Thus, the mass defect corresponds to the conversion of mass into binding energy that holds the nucleus together.

Q25. The half-life of a radioactive substance is 10 days. The time required for the substance to decay to one-eighth of its initial amount is:

Explanation:
Option C is correct. The remaining fraction of a radioactive substance is given by (1/2)ⁿ, where n is the number of half-lives. Here, 1/8 = (1/2)³, so the substance undergoes 3 half-lives to decay to one-eighth of its initial amount. , Given half-life = 10 days , Time required = 3 × 10 = 30 days.

Q26. A monochromatic light of frequency 5.5 × 10¹⁴ Hz is incident on a metal surface having work function 2.2 eV. The maximum kinetic energy of emitted electrons is:

Explanation:
Option A is correct. Energy of photon: E = hν = (6.6 × 10⁻³⁴) × (5.5 × 10¹⁴) = 3.63 × 10⁻¹⁹ J , Convert into eV: E = (3.63 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) ≈ 2.27 eV , Maximum kinetic energy: K.E = E − φ = 2.27 − 2.2 = 0.07 eV ≈ 0.08 eV , Always subtract work function from photon energy to get maximum kinetic energy.

Q27. The stopping potential for a metal is 2 V when light of frequency 8 × 10¹⁴ Hz is incident. What will be the stopping potential if frequency is doubled?

Explanation:
Option D is correct. Photoelectric equation: eV₀ = hν − φ , For first frequency ν: eV₁ = hν − φ ...(1) , For doubled frequency (2ν): eV₂ = h(2ν) − φ = 2hν − φ ...(2) , Subtract (1) from (2): e(V₂ − V₁) = hν , So, V₂ = V₁ + (hν / e) , Since hν/e is unknown (φ is unknown), we cannot find exact V₂. Therefore, the stopping potential cannot be determined uniquely.

Q28. Binding energy per nucleon is maximum for:

Explanation:
Option B is correct. Binding energy per nucleon increases with mass number up to the iron region (A ≈ 56) and then decreases. Nuclei around iron (Fe) have the highest binding energy per nucleon, making them the most stable. Therefore, iron has the maximum binding energy per nucleon among the given options.

Q29. In alpha decay, the atomic number and mass number change as:

Explanation:
Option A is correct. An alpha particle is a helium nucleus (²₄He), which contains 2 protons and 2 neutrons. When a nucleus emits an alpha particle: - Atomic number (Z) decreases by 2 (loss of 2 protons) , - Mass number (A) decreases by 4 (loss of 2 protons + 2 neutrons) , Hence, the change is Z − 2 and A − 4.

Q30. Radioactive decay is independent of:

Explanation:
Option D is correct. Radioactive decay is a nuclear process that depends only on the properties of the nucleus. It is not affected by external factors such as temperature, pressure, or chemical state. Therefore, radioactive decay is independent of all of these.

Q31. In β-decay, atomic number:

Explanation:
Option D is correct. There are two types of beta decay: β⁻ decay: a neutron converts into a proton, so atomic number (Z) increases by 1. , β⁺ decay: a proton converts into a neutron, so atomic number (Z) decreases by 1. , Hence, depending on the type of β-decay, the atomic number either increases or decreases by 1.

Q32. Nuclear fusion requires:

Explanation:
Option B is correct. Positively charged nuclei repel each other due to electrostatic (Coulomb) force. A very high temperature is required to provide enough kinetic energy to overcome this repulsion and bring the nuclei close enough for fusion to occur. Hence, nuclear fusion requires high temperature.

Q33. Energy released in nuclear fission is due to:

Explanation:
Option B is correct. In nuclear fission, the total mass of the products is slightly less than the mass of the original nucleus. This missing mass is called mass defect, which is converted into energy according to Einstein’s equation: E = Δmc² , Hence, energy released in nuclear fission is due to mass defect.

Q34. Heavy stable nuclei generally have:

Explanation:
Option B is correct. In heavy nuclei, the number of neutrons is greater than the number of protons. Neutrons help increase nuclear binding without adding electrostatic repulsion, thereby compensating for the strong repulsive force between protons. Hence, stable heavy nuclei have more neutrons than protons.

Q35. SI unit of radioactivity is:

Explanation:
Option B is correct. The SI unit of radioactivity is the becquerel (Bq). 1 becquerel is defined as 1 disintegration (decay) per second. , 1 Bq = 1 decay s⁻¹ , Hence, becquerel is the SI unit of radioactivity.

Q36. At absolute zero temperature, an intrinsic semiconductor behaves as:

Explanation:
Option C is correct. At absolute zero (0 K), an intrinsic semiconductor has no thermal energy to excite electrons from the valence band to the conduction band. As a result, there are no free charge carriers (electrons or holes), so it behaves like an insulator.

Q37. An n-type semiconductor is obtained by doping silicon with:

Explanation:
Option D is correct. Pentavalent impurity atoms (such as phosphorus or arsenic) have five valence electrons. When doped into silicon, four electrons form covalent bonds with neighboring silicon atoms, and the fifth electron becomes a free charge carrier. This increases the number of free electrons, making it an n-type semiconductor.

Q38. In a p-type semiconductor, the majority charge carriers are:

Explanation:
Option B is correct. A p-type semiconductor is formed by doping a pure semiconductor with trivalent impurities (such as boron). These impurities create vacancies called holes, which act as positive charge carriers. Hence, holes are the majority carriers, while electrons are the minority carriers.

Q39. The potential barrier in a p–n junction is formed due to:

Explanation:
Option D is correct. When a p–n junction is formed, charge carriers diffuse across the junction due to concentration difference. This leads to recombination of electrons and holes near the junction, leaving behind charged ions. These ions create an electric field, forming the depletion region and the potential barrier.

Q40. When a p–n junction diode is forward biased:

Explanation:
Option B is correct. In forward bias, the p-side is connected to the positive terminal and the n-side to the negative terminal of the battery. This reduces the width of the depletion region and lowers the potential barrier, allowing charge carriers to cross the junction easily and resulting in current flow.

Q41. In a p–n junction diode under reverse bias, the current is:

Explanation:
Option B is correct. In reverse bias, the depletion region widens and the potential barrier increases, preventing the flow of majority charge carriers. Only a small current due to minority carriers flows, known as reverse saturation current, which is very low but not zero.

Q42. A Zener diode is primarily used for:

Explanation:
Option C is correct. A Zener diode is designed to operate in the reverse breakdown region, where it maintains a nearly constant voltage across it despite changes in current. This property makes it suitable for voltage regulation.

Q43. A transistor, when suitably biased, can perform different functions depending on the region of operation. In the active region, it is capable of increasing the amplitude of input signals, while in cutoff and saturation regions, it can control the flow of current similar to an on-off device. Based on these characteristics, a transistor can be used as:

Explanation:
Option C is correct. A transistor acts as an amplifier when operated in the active region, where it increases the strength of input signals. It functions as a switch when operated in cutoff (OFF state) and saturation (ON state) regions, allowing or blocking current flow.

Q44. The output of an AND logic gate is 1 only when:

Explanation:
Option B is correct. An AND gate produces a high output (1) only when all its inputs are high (1). If any one of the inputs is 0, the output becomes 0.

Q45. The electrical conductivity of an intrinsic semiconductor increases with rise in temperature because:

Explanation:
Option B is correct. With increase in temperature, more covalent bonds break due to thermal energy, generating additional electron–hole pairs. This increases the number of charge carriers, thereby increasing the electrical conductivity of the semiconductor.

Test Summary