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When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.

Sagot :

Answer:

Explanation:

energy of photon having wavelength of 400 nm = 1237.5/400 eV

= 3.1 eV.

Maximum kinetic energy of photoelectrons = 1.1 eV .

Threshold energy Ф = 3.1 - 1.1 = 2 eV .

energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .

Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy

= 4 - 2 = 2 eV .

Required kinetic energy K₀= 2 eV.

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