The work function for caesium atom is 1.9 eV. Calculate
(a) the threshold wavelength and
(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
A) Work function of caesium (WO) = hvo
Therefore vo = WO/h = 1.9 x1.602 x10-19 / 6.626x10-34
= 4.59x1014/sec
B) λo =c/vo = 3x108 / 4.59x1014 = 6.54x10-7m
C) K.E of ejected electron = h(v-vo) = hc(1/λ – 1/ λo)
=(6.626x3x10-26) (1/500x10-9 – 1/654x10-9)
=(6.626x3x10-26) / 10-9(154/500x654)
= 9.36x10-20J
k.E = 1mv2/2 = 9.36x10-20J
=9.1x10-31/2 = 9.36x10-20J
Or
v2 = 20.55x1010m2s-2
Or
v = 4.53x105ms-1
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(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
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(g) H2S2O7
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(iii) used to determine pressure volume work
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(iii) molecular formula.
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Part 1st ka ans second wala h and vice versa
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