Question

-When light of wavelength λ illuminates the surface of Metal 1, the stopping voltage is V. In terms of V, what will be the stopping voltage if the same wavelength is used to illuminate the surface of Metal 2? The work function of Metal 1 is 5.7 eV and the work function of Metal 2 is 2.1 eV.

1-V – 3.6 volts

2-3.6 volts – V

3-2.1 volts – V

4-V + 3.6 volts

5-5.7 volts – V

Answer 1

--(1)

---(2)

from (1)

substituting the above result in (2), we get,

OPTION 4) is the right answer