When a metallic surface is illuminated with radiation of wavelength '$$\lambda$$', the stopping potential is '$$\mathrm{V}$$'. If the same surface is illuminated with radiation of wavelength '$$2 \lambda$$', the stopping potential is '$$\left(\frac{\mathrm{v}}{4}\right)$$'. The threshold wavelength for the metallic surface is

A metal surface of work function $$1 \cdot 13 \mathrm{~eV}$$ is irradiated with light of wavelength $$310 \mathrm{~nm}$$. The retarding potential required to stop the escape of photoelectrons is [Take $$\frac{\mathrm{hc}}{\mathrm{e}}=1240 \times 10^{-9} \mathrm{SI}$$ units]

The maximum kinetic energies of photoelectrons emitted are $$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ when lights of wavelengths $$\lambda_1$$ and $$\lambda_2$$ are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$ then

If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with electrons change?