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3 years ago

The work function for potassium is 2.24 eV. If potassium metal is illuminated with light of wavelength 350 nm , find:

1 answer:

8 0

Photo electric effect.Some guidance notes which may help ...
The energy of a photon of wavelength lambda is E=hc/lambda where h is Plank's constant, c is the speed of light, and lambda is the wavelength.The units of eV (electron volts) and the value of Planck's constant probably have to be determined. Calculate E=hc/lambda as hx3x10^8/(350x10^-9) and compare that to 2.24eV (expressed in consistent units which could be Joules)If the 350nm photon energy is higher than 2.24eV, then the difference should be the kinetic energy of the ejected photo electron.
if a photon has equal to or less than 2.24 eV work function, then it won't be ejected. That wavelength can be found by equating the work function to the photon energy using the equation E=hc/lambda2.24eV=hx3x10^8/(lambda_cut_off)

If the incoming light photon has an energy less than the work function of the metal, then the metal will not emit a photon.
This is the photo-electric effect, explained by Einstein who got a Nobel prize for this, and also for relativity. The point was, it seems, that it didn't matter how intense the light source was, it was the energy of photons within the light source, expressed by their wavelength, that mattered.
If you want any further information please ask.

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The given data is as follows.

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Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and $v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get: $v_{1} =0.30661(m^{3}/Kg)$ . Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get $v_{2}=0.23275(m^{3}/Kg)$ , knowing that it's a isobaric process we can find the compression's work as: $W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and: $W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric: $W_{2}=zero$ ,now $W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$ . Finally to get the temperarure at the final state in part (b) we get: $v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare $v_{f}$ and $v_{g}$ at the saturated water table and find the following: $v_{f}=0.001093(m^{3}/Kg)$ , so we know that the final state phase is a satured mixture and we get the temperature at the final state as: $T_{2} =T_{sat} =151.83^{0}C$ .

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