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Einstein's photoelectric equation with one reading: stopping potential, work function, kinetic energy or wavelength

This type of question, asked 5 times: 26 Jun 2022, 6 Apr 2024, 9 Apr 2024, 23 Jan 2025, 5 Apr 2026 S1. All 6 questions below, oldest first.

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  1. Question 1

    26 Jun 2022

    120
    The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630Å6630\text{Å} is 0.42 V0.42\text{ }V. If the threshold frequency is x×1013/sx \times 10^{13}/s, where xx is (nearest integer). (Given, speed light =3×108 m/s= 3 \times10^{8}\text{ }m/s, Planck's constant =6.63×10−34Js= 6.63 \times10^{- 34}Js )

  2. Question 2

    6 Apr 2024

    103
    When UV light of wavelength 300 nm300\text{ }nm is incident on the metal surface having work function 2.13eV2.13eV, electron emission takes place. The stopping potential is : (Given hc =1240eVnm= 1240eVnm )

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  3. Question 3

    6 Apr 2024

    20
    In photoelectric experiment energy of 2.48eV2.48eV irradiates a photo sensitive material. The stopping potential was measured to be 0.5 V0.5\text{ }V. Work function of the photo sensitive material is :

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  4. Question 4

    9 Apr 2024

    105
    UV light of 4.13eV4.13eV is incident on a photosensitive metal surface having work function 3.13eV3.13eV. The maximum kinetic energy of ejected photoelectrons will be:

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  5. Question 5

    23 Jan 2025

    81
    In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave? (Given hc =1242eVnm= 1242eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

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  6. Question 6

    5 Apr 2026 S1

    16
    Light source having wavelength 331 nm is used to generate photoelectrons whose stopping potential is 0.2 V . The work function of the used metal in the experiment is α×10−19 J\alpha \times10^{- 19}\text{ }J. The value of α\alpha is____\_\_\_\_ . ( h=6.62×10−34 J s,e=1.6×10−19Ch = 6.62 \times10^{- 34}\text{ }J\text{ }s,e = 1.6 \times10^{- 19}C and c =3×108 m/s= 3 \times10^{8}\text{ }m/s )

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