λhc=ϕ+eV ...........(1)
3λhc=ϕ+4eV ---------(2)
from (1) & (2)
λhc(1−31)=43ev
λhc32=43eV
eV=98λhc
λhc=ϕ+98λhc
ϕ=9λhc=λthhc
λth=9λ
∴k=9
When radiation of wavelength A is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3A, the stopping potential is 4V. If the threshold wavelength for the metallic surface is nλ then value of n will be :
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