(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 = 3×108 ms–1, e = 1.6×10–19 coulomb)
(1) 2.48 eV
(2) 0.41 eV
(3) 2.07 eV
(4) 0.82 eV
(2) A photon of energy ‘E’ejects a photo electron from a metal surface whose work function is W0. If this electron enters into a magnetic field of induction ‘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) √[2m(E – W0)/eB]
(2) √[2m(E – W0) eB]
(3) (1/mB)√[2e(E – W0)]
(4) (1/eB)√[2m(E – W0)]
3)Blue light (λ = 460 nm) is incident on a piece of potassium, for which the work function W= 2.20 eV. What is the maximum kinetic energy of the ejected photoelectrons?
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2 Answers
1K_{max} = h (f-f_{0})
f= frequency =\frac{c}{\lambda }
f_{0}= work \, \, function
f =\frac{3\times 10^8}{5\times 10^{-7}} =6\times 10^{14}\, \texttt{Hz}
K_{max} = 6.62\times 10^{-34} (6\times 10^{14}-3.31\times 10^{-19})
= 3.972\times 10^{-19}\textup{J}
K_{max} = 2.4825\, \textup{eV}
Option a]
302) The maximum kinetic energy that the electron can possess= E - W0
Assuming mass of electron to be m,
Momentum of ejected photo-electron = \sqrt{2m(E-W0)}
Radius 'r' described by the photo-eletcron in a magnetic field of induction B entering perpendicularly into it = mveB
= \frac{\sqrt{2m(E-W0)}}{eB} (Option (4) )
I guess in this question all other options are dimensionally incorrect too!
3) Maximum Kinetic Energy = E - φ = hcλ - φ = 1242eV-nm460nm - 2.20eV = 2.7eV - 2.2eV = 0.5eV
