The work function for mercury is equivalent to 435 kJ mol^-1...

The work function for mercury is equivalent to 435 kJ mol$^{-1}$ photons. (a) Can the photoelectric effect be obtained with mercury by using visible light? Explain. (b) What is the kinetic energy, in joules, of the ejected photoelectrons when light of 215 nm strikes a mercury surface? (c) What is the velocity, in m s$^{-1}$, of the ejected photoelectrons in part (b)?

Transcript

00:01 First of all by plank equation we know that e is equals to h nu so new is the frequency h is the plank constant putting the values i can also write h c divided by lambda here because new is equals to c divided by lambda right so now i will put the values here e is equal to so this is plank constant 6 .626 multiplying 10 to power minus 34 jule second multiplying speed of light that is 3 .0 multiplying 10 risk to power 8 meter second inverse okay this c is the speed of light divided by lambda that is wave that is 215 nanometer okay so since this is nanometer second inverse so i have to convert this nanometer to meter and we know that since one nanometer consists of 10 x to power minus 9 meter okay so multiplied by 10 to power minus 9 meter now solving this we will get energy so that will be 0 .0 924 multiplying 10 raise to power minus 17 solving all the powers and finally the final unit will be jule okay and shifting the decimal so i can write 9 .24 okay so now this is further multiplied by 10 raise to power minus 19 jule okay so this is the energy we have calculated and now we will apply the plank the photoelectric equation okay so by photo electric equation this was here plank equation so now we have relation that the energy of photon e so photon that is the that is the smallest packet form of energy the plank theory says that the energy radiation travels in the form of small packets of energy.

03:06 Okay, so these are the positive integral values.

03:09 So now that is equal to this phi, okay, and then we have less kinetic energy.

03:29 Okay.

03:37 Now if i want to release the electron out of the metal surface, so i have to give some three -should wavelength or three -should frequency of light, that is three -should frequency, that is the minimum frequency which is very important to release the electron outermost of the electron.

03:58 Okay, so that electron can come out of the metal surface, okay? so this energy that is called three -should energy.

04:08 And we also call that as work function.

04:11 Okay, so this phi here that is work function and kinetic energy so it says that now if the given incident energy of photon that is now it is remained extra so the extra part becomes converted into kinetic energy okay so i gave a incident energy of light to a metal then the electron comes out from the metal surface...

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