How Does Changing the Work Function Affect Photoelectron Emission?

[meltem]

Modern Physics, homework, please help:(

A metal surface with a work function of (W1=3eV) is illuminated with a monochromatic ultraviolet light of frequency v*1. The maximum speed of the ejected photo electrons was found to be v1=0.6x10^6m/s.

a) compute the wavelength λ of the incident photons.
b) what is the stopping potential?
c) now the metal is in the photo tube is changed so the work function is W2=5.5eV. Everything else remains the same. Compute the maxiumum speed of the ejected photo electrons and discuss.
d) How would the magnitude of the photo current be changed if the intensity of the incident light were increased? Increases, decreases or does not change? ( answer these question for parts (a) and (c)).

 

[rock.freak667]

You need to post your attempt. What would your relevant equations be?

 

[meltem]

You need to post your attempt. What would your relevant equations be?

sorry, i didnt know this rule.. actually i knew formulas but i don't know where i should use them. Also this question is from my exam. I couldn't do it there:s

a ) A metal surface with work function φ = 3.0 eV
maximum speed ejected from photo electrons is v_1 = 0.6 *10^6 m/s

K.E _max = h f - φ
1/2 mv^2 = hf - φ
1/2 * 9.1 *10^-31 kg ( 0.6 *10^6 m/s)^2 = 6.625*10^-34 J.s * f - 3.0 * 1.6 *10^-19 J /eV
f = 9.71*10^-5 Hz
wavelength λ = c
= 3 *10^8 m/s / 9.71*10^-5 Hz
= 3.08*10^12 m

b ) stopping potential is K.E = q ΔV
1/2 * 9.1 *10^-31 kg ( 0.6 *10^6 m/s)^2 = 1.6 *10^-19 C * ΔV
ΔV = 1.02V

c) same formula with a
d) i don't know:s

am I right?

 

[gneill]

You might want to revisit your frequency calculation. The method and numbers used look okay, but your result seems to be out of whack; your value implies a period for the wave of nearly three hours!

For photoelectric emission on metals you'd expect the light to be in the hundreds of terahertz range.

 

[rwisz]

a) To compute the wavelength λ of the incident photons, we can use the equation E = hv, where E is the energy of the photon, h is Planck's constant (6.626x10^-34 J*s), and v is the frequency of the light. We know that the energy of the photon must be equal to the work function of the metal (W1 = 3eV) plus the kinetic energy of the ejected photoelectrons (K.E. = 1/2 mv^2). So we can set up the equation 3eV + 1/2 mv^2 = hv. We also know that the maximum speed of the ejected photoelectrons is v1 = 0.6x10^6m/s. Rearranging the equation, we get v = √(2(hv1-W1)/m). Plugging in the values, we get v = 3.08x10^15 Hz. To convert this to wavelength, we can use the equation c = λv, where c is the speed of light (3x10^8 m/s). Solving for λ, we get a wavelength of 9.74x10^-8 m or 97.4 nm.

b) The stopping potential is the minimum potential difference needed to stop the ejected photoelectrons. We can calculate this by using the equation K.E. = eV, where e is the charge of an electron (1.6x10^-19 C) and V is the stopping potential. Plugging in the values, we get V = K.E./e = (1/2)(m)(v1^2)/e = 1.8 V.

c) With the new work function of W2 = 5.5eV, the maximum speed of the ejected photoelectrons can be calculated using the same equation as in part (a). However, this time we need to use the new work function and the new maximum speed of the ejected photoelectrons (v2). So the equation becomes 5.5eV + 1/2 mv2^2 = hv. Rearranging, we get v2 = √(2(hv2-W2)/m). Plugging in the values, we get v2 = 2.45x10^15 Hz. Converting to wavelength using the equation from part (a), we get a wavelength of