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NEET-UG and JEE Main Chemistry · Class 11

Electromagnetic radiation and the photoelectric effect — NEET & JEE Questions

Electromagnetic radiation and the photoelectric effect is a topic in Atomic Structure, a Class 11 Chemistry chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 27 published NEET-UG and JEE Main questions on Electromagnetic radiation and the photoelectric effect, each with the marking scheme of the real paper.

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Electromagnetic radiation and the photoelectric effect MCQs with solutions

10 of the 27 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    The energy of a photon is given by

    • A. E=hνE=h\nu
    • B. E=h/νE=h/\nu
    • C. E=ν/hE=\nu/h
    • D. E=h+νE=h+\nu
  2. Question 2 · difficulty L2 · understanding

    Consider two radiations of wavelengths λ1=2000Ao\lambda_1=2000\mathop {\rm{A}}\limits^{\rm{o}} λ2=6000Ao\lambda_2=6000 \mathop {\rm{A}}\limits^{\rm{o}} The ratio of the energies of these two radiations (E1E2)\left(\frac{E_1}{E_2}\right) is ____\_\_\_\_ (Nearest integer).

  3. Question 3 · difficulty L2 · understanding

    The surface of sodium metal is irradiated with radiation of wavelength xx nm. The kinetic energy of ejected electrons is 2.8×10−20 J2.8 \times 10^{-20}\ \text{J}. The work function of sodium is 2.32.3 eV. The value of xx is ________ ×102\times 10^2 nm. (Nearest integer) (Given: h=6.6×10−34 J sh=6.6 \times 10^{-34}\ \text{J s} ; 1 eV=1.6×10−19 J1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J} ; c=3.0×108 m s−1c = 3.0 \times 10^8\ \text{m s}^{-1})

  4. Question 4 · difficulty L2 · understanding

    A hypothetical electromagnetic wave is show below. The frequency of the wave is x×1019 Hz\mathrm{x} \times 10^{19} \mathrm{~Hz}. x=\mathrm{x}= _________ (nearest integer)

  5. Question 5 · difficulty L2 · understanding

    The ionization energy of sodium in  kJ mol−1\mathrm{~kJ} \mathrm{~mol}^{-1}, if electromagnetic radiation of wavelength 242 nm242 \mathrm{~nm} is just sufficient to ionize sodium atom is _______.

  6. Question 6 · difficulty L2 · understanding

    Values of work function (W0_0) for a few metals are given below Metal Li Na K Mg Cu Ag W0_0/eV 2.42 2.3 2.25 3.7 4.8 4.3 The number of metals which will show photoelectric effect when light of wavelength 400 nm400 \mathrm{~nm} falls on it is ___________ Given: h=6.6×10−34 J s\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} c=3×108 ms−1c=3 \times 10^{8} \mathrm{~ms}^{-1} e=1.6×10−19Ce=1.6 \times 10^{-19} \mathrm{C}

  7. Question 7 · difficulty L2 · understanding

    The number of incorrect statement/s about the black body from the following is __________ (A) Emit or absorb energy in the form of electromagnetic radiation. (B) Frequency distribution of the emitted radiation depends on temperature. (C) At a given temperature, intensity vs frequency curve passes through a maximum value. (D) The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature.

  8. Question 8 · difficulty L2 · understanding

    If the work function of a metal is 6.63 ×\times 10 −-19 J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer) [Given : h = 6.63 ×\times 10 −-34 J s, and c = 3 ×\times 10 8 m s −-1 ]

  9. Question 9 · difficulty L2 · understanding

    A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x ×\times 10 20 . The value of x is __________. [Given : h = 6.63 ×\times 10 −-34 Js and c = 3.0 ×\times 10 8 ms −-1 ]

  10. Question 10 · difficulty L2 · understanding

    The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x ×\times 10 13 . The value of x is _____________. (Nearest integer) (h = 6.63 ×\times 10 −-34 Js, c = 3.00 ×\times 10 8 ms −-1 )

Common mistakes in Electromagnetic radiation and the photoelectric effect

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Uses a square or inverse dependence.
  • Assumes intensity can compensate for frequency below threshold.
  • Assumes intensity controls individual photon energy.
  • Treats many low-energy photons as if their energies combine for one electron.

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