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QUESTION 47given material has work function of 1.76 eV: White lignt with wavel engths over ange from 374,5nmto 714nm is incident on this material What is the maximu...

Question

QUESTION 47given material has work function of 1.76 eV: White lignt with wavel engths over ange from 374,5nmto 714nm is incident on this material What is the maximum kinetic energy ofan ectron which can be ejected from tne material by this Ilght? (1 nm 1.71536349132176 ev '5594259012016 ev 1,09159813084112 ev 2.18319626168224 ev 0.77971295060080-

QUESTION 47 given material has work function of 1.76 eV: White lignt with wavel engths over ange from 374,5nmto 714nm is incident on this material What is the maximum kinetic energy ofan ectron which can be ejected from tne material by this Ilght? (1 nm 1.71536349132176 ev '5594259012016 ev 1,09159813084112 ev 2.18319626168224 ev 0.77971295060080-



Answers

Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.68 eV to be ejected from a metal whose work function is 2.75 eV. What will be the maximum kinetic energy (in eV)
with which this same radiation ejects electrons from another metal whose work function is 2.17 eV?

So we're going to use Equation 38 5 In order to find the maximum kinetic energy of the ejected electrons. We know that this would be equal to the planks constant multiplied by the frequency of the incident radiation and then minus the materials work function. And so this would be equal to 4.14 times 10 to the negative 15th electron volts. This would be multiplied by 3.0 times, 10 to the 15th hurts. And then this would be minus the materials work function, 2.3 electron volts, and we find that the maximum kinetic energy of the inducted of the ejected electrons would be approximately. So we're going to use the equation 38 5 In order to find the maximum kinetic energy of the ejected electrons. We know that this would be equal to the planks constant multiplied by the frequency of the incident radiation and then march 10 0.1 electron volts. This would be our final answer. That is the end of the solution. Thank you. For what

Belongs to the modern physics in which we have a photo cathode that is emitting electrons with maximum kinetic energy E it is equals to 0.85 electron volt. When illuminated with wavelength lambda equals to 4 30 nanometer violet light. So for the part we have to determine village exact electron under equivalent lambda read. It is equal to 6 33 nanometer. Okay, so for this question, first of all, calculating the work function of the material. So we can write from the kinetic photoelectric effect equation so we can write that At Sea Belinda. This will be equals two kinetic energy maximum plus five. Okay, so from here we get five. This is equals two at symbol lambda minus kinetic energy. Okay, so maxim kinetic energy, it can be simply return its kinetic energy. So substituting values so we get five. It is equal to X, which is Planck's constant 6.626 molecular way 10 to the power -34, replaced by C. Is the speed of light. So 3 to 10 to the power eight divided by incident wavelength which is 4 30 nanometer. So 10 to the power minus nine minus kinetic energy which is this value. So 0.85 particular by 1.6 to 10 to the power minus 19. And this value comes out to be in jewel. So from here we get five. This is one function of the material, so it is equal to 3.26 molecular by 10 to the power minus 19 jewel. Okay, now 40 part A. We have this equivalent so calculating energy for this equivalent solar energy for the red light, it is equals two at Ciba lambda Red. So substituting value. So we get 6.626 molecular by 10 to the power minus 34 molecular by three and 2. 10 to the power eight m per second divided by lambda Red which is 6 33 particular by 10 to the power -9 m from here we get energy for the red light. It is equal to 3.14 particular by 10 to the power minus 19 jewel. So here we can see that we're function. Fight it is greater than the energy of incident light that is red light. So from the theory of photo electrons, we can see here that the photo electrons will not be omitted or we can say that there will be no ejection of photo flecktones when red light is used. So the answer for this question will be we can say that No. Okay, the answer will be here. No. The photo flecktones under red light will not be admitted or rejected. Now moving to the part B. So we have to calculate here the three short wavelength for this material that is linda. Not So linda, not, it is equal to at sea by five, So substituting their lives. So edges against 6.626 multiply by 10 to the power minus 34 june 2nd C. Is the speed of light, so 3 to 10 to the power eight m per second, divided by five, which is three point 26 molecular weight 10 to the power minus 19 jewels. So from here, three should be linda. Not. It is equal to 609.890 m. So this becomes the answer for the part B of the problem. Okay.

Yeah. Hi, everyone. Here it is. Given a light of frequency 1.88 10 to the power 15 Hertz incident on blacking up Having the world function 6.35 electoral gold Maximum kinetic energy of electron emptied by platinum We have to calculate. So it is proving mhm edge into F minus one function of platinum 6.63 10 to the power by this 30 ft frequency is 1.88 10. To leave our 15, this will be ensured. So you have to convert an electron volt. Uh huh. Minus 6.35 election vote. So it is to be equal to 1.44 electron volt. Mhm. Mm mm. Now, for iron, what conscience is mhm? 4.5. Reckon boat. So maximum kind of tech energy empirically iron surface. Yeah. 6.63 10 to the power minus 30 ft and to 1.88 10 to leave. Power 15, but 0.6 10 to the power minus 19 minus 4.5. So maximum kinda technology of ejected out. Electronics by iron surface. Yes. Mhm 3.29 Reckon Butch heads. You can compare it kind of technology of maximum contemporary maximum kinda technology of ejected or electron by iron surface. Its debt of Latin observe its no second part, that's all. Thanks for watching is.

In the execution we have given the work function office certain materials and we have to Calculate its three short frequency. So first of all, I am writing the work function, which is 5.8 electron world. Now we have to calculate the threshold frequency. So threshold frequency is given by F equals two E device edge. Now put the values of E and edge So f equals to this is 5.8 device, This is four point 1413 and 2 10 to the power minus 15. After solving this, we get the venue of frequency F. That is 1.4 and 2 10 to the Power 15. Nice. So this is the required to result frequency of the material. Thank you.


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