The unsolved question from my latest examination .. can YOU solve it? (1 Viewer)

[Fizzy_Cyst]

Good luck

View attachment Question 26.pdf

tee hee

I will give 200 of my OW SR to the first person to solve it

 

[calamebe]

a)

5V between the electodes, Energy = Vq = 5x1.602x10^(-19) = 8.01 x 10^(-19).
Hence, (mv^2)/2 = 8.01 x 10^(-19). v = 1,326,159 m/s = 1.33 x 10^6 m/s. This is the minimum. So the maximum speed will be this plus the speed of the electrons which are emitted by the highest frequency of light. Lowest wavelength of visible: 390nm -> 7.692 x 10^14 Hz -> 5.097 x 10^-19 J = 3.18 eV of energy. Hence, 1/2 mv^2 = 3.18 - 2.36 = 0.822 eV = 1.316 x 10^-19 J. v=537,577.5 m/s. Hence, the range is:

1.33 x 10^6 m/s <= v <= 1.87 x 10^6 m/s

EDIT: Max: mv^2/2 = hf – 2.36x1.602x10^-19 + 5x1.602x10^-19 -> v=1430975 m/s which is just obtained by substituting in the frequency (which corresponds to 390 nm), Plank's constant, and the mass of the electron, and then square rooting the velocity.

1.33 x 10^6 m/s <= v <= 1.43 x 10^6 m/s


b)

So F=Eq is the force accelerating the electrons.

So x = ut + 1/2 at^2 = ut + 1/2 F/m t^2. Substitute in x= 0.15, t = 200x10^-9, F=Eq = Vq/d = 5q/0.15 and the mass of electron to solve for the initial velocity.

0.15 = 2x10^(-7) u + 0.1172, therefore u = 163,766.6 m/s. hf - 2.36 = 1/2 mu^2, hf = 2.36 + 1/2 mu^2 = 3.903 x 10^(-19). Hence f =5.89 x 10^14, and λ = 3x10^8/f =5.09 x 10^7.

Btw I'm not sure how correct this is.

 

[pikachu975]

Isn't the range for visible light like 400-700 nm or 410 to 656 nm???

 

[Fizzy_Cyst]

a)

5V between the electodes, Energy = Vq = 5x1.602x10^(-19) = 8.01 x 10^(-19).
Hence, (mv^2)/2 = 8.01 x 10^(-19). v = 1,326,159 m/s = 1.33 x 10^6 m/s. This is the minimum. So the maximum speed will be this plus the speed of the electrons which are emitted by the highest frequency of light. Lowest wavelength of visible: 390nm -> 7.692 x 10^14 Hz -> 5.097 x 10^-19 J = 3.18 eV of energy. Hence, 1/2 mv^2 = 3.18 - 2.36 = 0.822 eV = 1.316 x 10^-19 J. v=537,577.5 m/s. Hence, the range is:

1.33 x 10^6 m/s <= v <= 1.87 x 10^6 m/s

b)

So F=Eq is the force accelerating the electrons.

So x = ut + 1/2 at^2 = ut + 1/2 F/m t^2. Substitute in x= 0.15, t = 200x10^-9, F=Eq = Vq/d = 5q/0.15 and the mass of electron to solve for the initial velocity.

0.15 = 2x10^(-7) u + 0.1172, therefore u = 163,766.6 m/s. hf - 2.36 = 1/2 mu^2, hf = 2.36 + 1/2 mu^2 = 3.903 x 10^(-19). Hence f =5.89 x 10^14, and λ = 3x10^8/f =5.09 x 10^7.

Btw I'm not sure how correct this is.

Everything is correct EXCEPT your max speed in part (a)! Niiiiiiiice!!

 

[Fizzy_Cyst]

View attachment 34215
I feel like this is the solution to 26a.
EDIT - Does the solution to part b involve a quadratic with very bad coefficients?

Correct approach, but 700nm does not cause the PE Effect. So your max value is correct, but min value is not