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Why are the photoelectric work functions...

Why are the photoelectric work functions different for differet metals?

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Electrons in a metal are delocalised and move freely throughout the crystal lattice of the metal. Hence each electron has to do some work to overcome the force of attraction of the metal ions.the amount of energy requried to eject the electrons (known as work function) depends on the metal. hence, different metals have different work functions.
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What do you mean by photoelectric work function?

When light of wavelength 4041 Å falls on a metal surface, the maximum velocity of the emitted photoelectrons becomes twice the maximum velocity when another light of wavelength 5893 Å falls on the same surface. What is the photoelectric work function of that metal?

A metal surface does not emit any electron if the energy of incident radiation is less than the photoelectric work function of the metal - justify the statement mathematically.

Write down the relation between threshold frequency and photoelectric work function for a metal.

Write down Einstein's photoelectric equation and mention the symbols used. The photoelectric threshold wavelength for a certain metal is 400 nm. Find the maximum kinetic energy of the emitted electrons from the metal surface by ultraviolet light of wavelength 200 nm. Given, h= 6.63 xx 10^(-34) J.s

Write down Einstein's photoelectric equation and mention the symbols used. The photoelectric threshold wavelength for a certain metal as 400mm. Find the maximum kinetic enrgy of the emitted electrons from the metal surface by ultraviolet light of wavelength 200nm. Given h=6.63xx10^34

The photoelectric work function of a metal is 1.2 eV. What will be the stopping potential for a radiation of wavelength 5500 Å incident on its surface?

Ratio of the work functions of two metal surface is 1 : 2 . If threshold wavelength of photoelectric effect for the 1st metal is 6000Å , what is the corresponding value for the 2nd metal surface?

If the photoelectric threshold wavelength of metallic silver is 3500 Å , and UV light of wavelength 2000 Å falls on it, find (i) the maximum kinetic energy of the photoelectrons, (ii) the maximum velocity of the photoelectrons, (iii) the value of work function of silver in joule

Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C The threshold wavelength of photoelectric effect for a metal surface is 4600 Å . Work function of the metal (in eV) is

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