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In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy (in eV) of emitted electrons will be: Given E(ineV)=1237/λ(innm)
    Correct answer is '1.5'. Can you explain this answer?
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    In a photoelectric effect experiment the threshold wavelength of light...
    To find the maximum kinetic energy of emitted electrons, we can use the equation:

    E = hc/λ - hc/λ₀

    where:
    E is the maximum kinetic energy of emitted electrons,
    h is Planck's constant (6.626 x 10^-34 J·s),
    c is the speed of light (3.00 x 10^8 m/s),
    λ is the wavelength of the incident light, and
    λ₀ is the threshold wavelength.

    Converting the given values to the appropriate units:
    λ = 260 nm = 260 x 10^-9 m
    λ₀ = 380 nm = 380 x 10^-9 m

    Plugging these values into the equation, we get:

    E = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (260 x 10^-9 m) - (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (380 x 10^-9 m)

    Calculating this expression, we find:

    E ≈ 2.42 eV

    Therefore, the maximum kinetic energy of emitted electrons is approximately 2.42 eV.
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    In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy (in eV) of emitted electrons will be: GivenE(ineV)=1237/λ(innm)Correct answer is '1.5'. Can you explain this answer?
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