A back potential of 0.38 V and and 1.85 V stop photo currents due to light wave

Question

A back potential of 0.38 V and and 1.85 V stop photo currents due to light wave lengths (lambda) 770 mm and 420 mm respectively. A) based on the information given, determine "h", Planck's constant. B) determine the work function of the material constituting the photo-tubes cathode C) determine the wavelength that produces a maximum kinetic energy of 0.75 eV.
Okay so I am a little confused on what to do here. I know Planck's constant is equal to 6.626 x 10^-39 J*s. Are we proving this within the values we are given or are we just writing an expression in terms of h? Kmax=hf-phi Kmax=(hc/lambda) - phi Which will lead to: .38= h(3 x 10^8 / 780 nm) - phi And 1.81= h(3 x 10^8/ 420 nm) - phi Respectively Is this the solution to part A or do we actual solve for h? I don't see how we can do this since we are not given phi. A little clarification would be most appreciated. I know the next two parts rely on part A so if anyone can help me here with all of them that would be great. Thanks guys. A back potential of 0.38 V and and 1.85 V stop photo currents due to light wave lengths (lambda) 770 mm and 420 mm respectively. A) based on the information given, determine "h", Planck's constant. B) determine the work function of the material constituting the photo-tubes cathode C) determine the wavelength that produces a maximum kinetic energy of 0.75 eV.
Okay so I am a little confused on what to do here. I know Planck's constant is equal to 6.626 x 10^-39 J*s. Are we proving this within the values we are given or are we just writing an expression in terms of h? Kmax=hf-phi Kmax=(hc/lambda) - phi Which will lead to: .38= h(3 x 10^8 / 780 nm) - phi And 1.81= h(3 x 10^8/ 420 nm) - phi Respectively Is this the solution to part A or do we actual solve for h? I don't see how we can do this since we are not given phi. A little clarification would be most appreciated. I know the next two parts rely on part A so if anyone can help me here with all of them that would be great. Thanks guys. A back potential of 0.38 V and and 1.85 V stop photo currents due to light wave lengths (lambda) 770 mm and 420 mm respectively. A) based on the information given, determine "h", Planck's constant. B) determine the work function of the material constituting the photo-tubes cathode C) determine the wavelength that produces a maximum kinetic energy of 0.75 eV.
Okay so I am a little confused on what to do here. I know Planck's constant is equal to 6.626 x 10^-39 J*s. Are we proving this within the values we are given or are we just writing an expression in terms of h? Kmax=hf-phi Kmax=(hc/lambda) - phi Which will lead to: .38= h(3 x 10^8 / 780 nm) - phi And 1.81= h(3 x 10^8/ 420 nm) - phi Respectively Is this the solution to part A or do we actual solve for h? I don't see how we can do this since we are not given phi. A little clarification would be most appreciated. I know the next two parts rely on part A so if anyone can help me here with all of them that would be great. Thanks guys.

Explanation / Answer

A) in this case you get two equations ,

.38= h(3 x 10^8 / 770 nm) - phi

=>0.38 = h * 3.896x1014 - phi ...........................(1)

and

1.85= h(3 x 10^8/ 420 nm) - phi

=> 1.85 = h* (7.14x1014) -phi .............................(2)

now these are two equations with two unkowns ,

subtract , (1) from (2)

=> 1.85-0.38 = h*( 7.14x1014- 3.896x1014) - phi + phi

( you can see that the phi term will cancel out)

h = 4.53144 x10-15 eV

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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