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Lesechka [4]
3 years ago
8

When a light of wavelength 570nm falls on the surface of potassium metal, electrons are ejected with a velocity of 6.4x104 m/s.

What is the minimum energy required to remove an electron from potassium metal?
Chemistry
1 answer:
Anarel [89]3 years ago
5 0

Answer:

3 x 10 m/s. = = 188 m. 1600 x 10 Hz λ. 8 max. 3. 3 x 10. = = 566 m. 530 x 10. FM λ. 8 ... (a) The equation relating and is c = where c is the speed of light = 3.00 x 10 ... Potassium metal can be used as the active surface in a photodiode because ... wavelength light (in nm) with quanta of sufficient energy to eject electrons from a.

Explanation:

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