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).(Give…

Question

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^-34Js)

Accepted Answer

Correct answer: 35. (35) Stopping potential V0 = 0.42 V = 6630Å E = + eV0 E: energy of incident photon V0: Stopping potential = E - eV0 =E-eV_0 E= 124006630 eV=1.87 eV =(1.87-0.42)=1.45 eV =hv_0 ;v_0 : threshold frequency 1.45 x 1.6 x 10-19 = 6.63 x 10-34x v0 v0 = 0.35 x 1015 = 35 x 1013 sec^-1 = 35

$The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630Å is 0.42 V. If the threshold frequency is x \times 1 0^{13} /s, where x is______(nearest integer).(Given, speed light = 3 x 1 0^{8} m/s, Planck's constant = 6.63 x 1 0^{- 34} Js)$

Solution

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