The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630Å is 0.42 V. If the threshold frequency is x × × 10 13 /s, where x is______(nearest integer).(Given, speed light = 3 x 10 8 m/s, Planck's constant = 6.63 x 10 -34 Js)
Text Solution
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CHECK THE SOLUTION.
(35)
Stopping potential V 0 = 0.42 V
= 6630 Å
E = 
+ eV 0
E: energy of incident photon
V 0 : Stopping potential
= E - eV 0
threshold frequency
1.45 x 1.6 x 10 -19 = 6.63 x 10 -34 x v 0
v 0 = 0.35 x 10 15
= 35 x 10 13 sec -1
= 35
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