The difference between threshold wavelengths for two metal surfaces $$\mathrm{A}$$ and $$\mathrm{B}$$ having work function $$\phi_{A}=9 ~\mathrm{eV}$$ and $$\phi_{B}=4 \cdot 5 ~\mathrm{eV}$$ in $$\mathrm{nm}$$ is:

$$\{$$ Given, hc $$=1242 ~\mathrm{eV} \mathrm{nm}\}$$

A proton and an $$\alpha$$-particle are accelerated from rest by $$2 \mathrm{~V}$$ and $$4 \mathrm{~V}$$ potentials, respectively. The ratio of their de-Broglie wavelength is :

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :

(Assume $$m_{p}=m_{e} \times 1849$$ )

A metallic surface is illuminated with radiation of wavelength $$\lambda$$, the stopping potential is $$V_{0}$$. If the same surface is illuminated with radiation of wavelength $$2 \lambda$$. the stopping potential becomes $$\frac{V_{o}}{4}$$. The threshold wavelength for this metallic surface will be