When light of wavelength 366 nm falls on a potassium surface, electrons are emit

Question

When light of wavelength 366 nm falls on a potassium surface, electrons are emitted that have a maximum kinetic energy of 1.91 eV. The speed of light is 3 × 108 m/s andPlanck’s constant is 6.63 × 1034 J · s .
The work function is 1.4865 eV
What is the cutoff wavelength of potassium? Answer in units of nm.
What is the threshold frequency for potas- sium? Answer in units of Hz.
I cannot figure out how to do the second two partsThanks.
When light of wavelength 366 nm falls on a potassium surface, electrons are emitted that have a maximum kinetic energy of 1.91 eV. The speed of light is 3 × 108 m/s andPlanck’s constant is 6.63 × 1034 J · s .
The work function is 1.4865 eV
What is the cutoff wavelength of potassium? Answer in units of nm.
What is the threshold frequency for potas- sium? Answer in units of Hz.
I cannot figure out how to do the second two partsThanks.

Explanation / Answer

The cuttoff wavelength of potassium is             = hc/             = (6.63*10-34J.s)(3*108m/s)/(1.4865eV(1.6*10-19J/1eV))                =---- m The threshold frequency of for potassium is                f =/h                   = (1.4865eV(1.6*10-19J/1eV))/(6.63*10-34J.s)                   =---- Hz Simplifying we get the answer.

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