When light of wavelength 271 nm is incident on potassium, the emitted electrons

Question

When light of wavelength 271 nm is incident on potassium, the emitted electrons have maximum kinetic energy of 4.53 eV

A) what is the energy of an incident photon? the value of hc is 1240 eV*nm? not need i already have the answer -- h=E/f 1240/271=4.5756eV

what i need help with is the next three

B) what is the work function for potassium? Answer in unites of eV

C) what would be the maximum kinetic energy of the electrons if the incident light had wavelength of 397nm? Answer in unites of eV

D) what is the threshold wavelength for the photoelectric effect with potassium? Answer in unites of nm

Explanation / Answer

A) Energy(in eV)=1240/Lambda(in nm)=1240/271=4.5756 eV

B) Work function = incident energy-K.E max= 4.5756-4.53= 0.0456 eV

C) Incident energy= 1240/Lambda(in nm)=1240/397=3.1234 eV

==> Kmax= 3.1234-0.0456 eV= 3.0778 eV

D) threshold wavelength= 1240/workfunction= 1240/0.0456= 27192.98 nm

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.