In a photoelectric effect experiment, the work function is 2.5eV. In terms of el
ID: 1387017 • Letter: I
Question
In a photoelectric effect experiment, the work function is 2.5eV. In terms of electron current magnitude I0 (for stopping voltage set to zero) and of stopping voltage Vs: What is the result of increasing the intensity of the light of wavelength 656 nm illuminating the anode? I0 (increases, decreases, is zero) Vs (increases. decreases, doesn't exist because I0 = 0. What is the result of increasing the intensity of the light of wavelength 434 nm iluminating the anode? I0 (increases, decreases, is zero) Vs (increases, decreases, doesn'texist because I0 = 0. What is the result of decreasing the wavelegth from 434 nm to 350nm. I0 (I'll give you this one: decreases slightly, since there are fewer photons for the same amount of energy or power.) Vs (increases, decreases, doesn't exist because I0 = 0.Explanation / Answer
work functiom Wo= 2.5*1.6*10^-19 = 4e-19 J
h*c/lamda = 4*10^-19
lamda = (6.625*10^-34*3*10^8)/(4*10^-19) = 496.8 nm
frequency f = (4*10^-19)/(6.625*10^-34) = 6.03*10^14 Hz
A) for lamda = 656 nm
f = (3*10^8)/(656*10^-9) = 4.57*10^14 Hz
Which is less than 6.03*10^14 Hz..hence elctrons are not ejected from metal surface
hence current Io is Zero
Vs doesnot exist because Io = 0
B) f = (3*10^8)/(434*10^-9) = 6.91*10^14 Hz
which is greater than 6.03*10^14 Hz
hence current increases
with frequency stopping potential Vs also increases
C) lamda = 350 nm
f = (3*10^8)/(350*10^-9)= 8.57*10^14 > 6.03*10^14
Io increases
Vo also increases
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