Questions on photoelectric effect are usually interesting. One or two questions in this section often find place in entrance examination question papers. I give below two questions on photoelectric effect which were included in the EAMCET (Engineering) 2009 question paper:

**(1)** The work function of a certain metal is 3.31×10^{–19} J. Then the maximum kinetic energy of photo electrons emitted by incident photons of wave length 5000 Ǻ is (given *h* = 6.62×10^{–34} Js, *c = *3×10^{8} ms^{–1}, *e = *1.6×10^{–19} coulomb)

(1) 2.48 eV

(2) 0.41 eV

(3) 2.07 eV

(4) 0.82 eV

It will be useful (as I have mentioned on earlier occasions) to remember that a photon of wave length 1000 Ǻ has energy of 12.4 eV. In other words, the product of the wavelength in Angstrom (Ǻ) and the energy in electron volt (eV) of any photon is 12400 units. Therefore, the energy of 5000 Ǻ photon is 12400/5000 = 2.48 eV.

[You can calculate the energy (*E*) of the photon in *joule* using the equation, *E = hc/λ* and then convert it into *electron volt* by dividing it by 1.6×10^{–19} as 1 eV = 1.6×10^{–19} joule. But you will have to spend too much of your valuable time for this].

The work function of the metal is 3.31×10^{–19} J = (3.31×10^{–19})/(1.6×10^{–19}) eV = 2.07 eV nearly.

The maximum kinetic energy of photoelectrons emitted = Energy of incident photon – Work function = 2.48 eV – 2.07 eV = 0.41 eV.

**(2) **A photon of energy ‘*E*’ejects a photo electron from a metal surface whose work function is *W*_{0}. If this electron enters into a magnetic field of *B*’ in a direction perpendicular to the field and describes a circular path of radius ‘*r*’, then the radius ‘*r*’ is given by (in the usual notation)

(1) √[2*m*(*E* – *W*_{0})/*eB*]* *

(2) √[2*m*(*E* – *W*_{0})* eB*]* *

(3) (1/*mB*)√[2*e*(*E* – *W*_{0})]

(4) (1/*eB*)√[2*m*(*E* – *W*_{0})]

The kinetic energy (*k*) of the photo electron is given by

*k = E *– *W*_{0}

The momentum *p *of the photo electron is related to its kinetic energy as

*k = p*^{2}/2*m,*

Therefore, momentum *p =*√(2*mk*) = √[2*m*(*E* – *W*_{0})]

The radius ‘*r*’* *of the circular path followed by the electron in the magnetic field is given by

*r* = *mv/qB *

[This is obtained by equating the magnetic force to the centripetal force:

*qvB = mv*^{2}/*r*]

Since *mv = p*, we have *r = p/qB = p/eB = *(1/*eB*)√[2*m*(*E* – *W*_{0})].

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