SELECT * FROM question_mgmt as q WHERE id=1300 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=43 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
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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