Recent content by Stevee

600nm.. [6.64x10^-34*3x10^8]/600x10^-9 = ∑ + 1.76x10^-19 unknown nm... [6.64x10^-34*3x10^8]/ lambda = ∑ + 1.76x10^-19

Oh ok, ofcourse haha, so from this we can work out the photon energy?

so it would be equivalent to the energy required to emit a photoelectron from the tube?

Thankyou, hmm it doesn't seem to be working out for me. So i used E=q*V E=(1.6x10^-19)*(1.10)=1.76x10^-19 into... E=h*c/lambda 1.76x10^-19=(6.63x10^-34)*(3x10^8)/lambda where lambda comes out as113x10^-8m. im doing something wrong

q=E/V? oh woww.. haha You use E=qV to find the energy then just sub it into E=h*c/lambda and solve for lambda right?..

Oh, sorry photoelectron not electron.. my mistake well I am not entirely sure, I am having a real blank at the moment... I am not sure on whether its possible to use the first wavelength (and stopping pot.) to calculate the work function to sub into the equation, but the frequency changes with...

So, the stopping potential would be equivalent to the available kinetic energy which is [Avail. Ek=E - sigma] right? Do i then work out the work function from the other provided wavelength? The stopping potential tells us the amount of voltage required to stop the movement of electrons, so into...

Homework Statement A stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 600 nm. Monochromatic light of a different wavelength is now shone on the tube, and the stopping potential is measured to be 1.1V. What is the wavelength of this...