Modern Physics 2 Question 6

6. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300nm to 400nm. The decrease in the stopping potential is close to hce=1240nmV

(Main 2019, 11 Jan II)

(a) 0.5V

(b) 2.0V

(c) 1.5V

(d) 1.0V

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Solution:

  1. Given, λ1=300nm

λ2=400nmhce=1240nm

Using Einstein equation for photoelectric effect,

E=hv=φ+eV0

(here, φ is work function of the metal and V0 is stopping potential)

For λ1 wavelength’s wave,

E1=hν1=φ+eV01 or hcλ1=φ+eV01 Similarly, hcλ2=φ+eV02

From Eqs. (ii) and (iii), we get

hc1λ11λ2=e(V01V02) or hce1λ11λ2=ΔV

By using given values,

ΔV=124013001400nmVnm=1240×11200VΔV=1.03.V1V



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