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Photoelectric threshold: minimum photon energy or maximum wavelength from the work function

This type of question, asked 3 times: 25 Jun 2022, 28 Jun 2022, 12 Apr 2023. All 3 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    121
    The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is: [Given: The threshold frequency of platinum is 1.3 ×1015 s−1\times10^{15}{\text{ }s}^{- 1} and h=6.6×10−34 J sh = 6.6 \times10^{- 34}\text{ }J\text{ }s.]

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  2. Question 2

    28 Jun 2022

    52
    If the work function of a metal is 6.63×10−19 J6.63 \times10^{- 19}\text{ }J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____ nm. (Nearest integer) [Given: h=6.63×10−34 J sh = 6.63 \times10^{- 34}\text{ }J\text{ }s, and c=3×108 m s−1c = 3 \times10^{8}\text{ }m{\text{ }s}^{- 1} ]

  3. Question 3

    12 Apr 2023

    60
    Values of work function (W0)\left( W_{0} \right) for a few metals are given below:
    MetalLiNaKMgCuAg
    W0/eVW_{0}/\mathrm{eV}2.422.32.253.74.84.3
    The number of metals which will show photoelectric effect when light of wavelength 400 nm400\text{ }nm falls on it is__ Given: h=6.6×10−34 J sh = 6.6 \times10^{- 34}\text{ }J\text{ }s
    c=3×108 m s−1c = 3 \times10^{8}\text{ }m{\text{ }s}^{- 1}
    e=1.6×10−19Ce = 1.6 \times10^{- 19}C