Proton $$(\mathrm{P})$$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $$\mathrm{m}_{\mathrm{p}}=1849 \mathrm{~m}_{\mathrm{e}}$$ ):

The work functions of Aluminium and Gold are $$4.1 ~\mathrm{eV}$$ and and $$5.1 ~\mathrm{eV}$$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

The kinetic energy of an electron, $$\alpha$$-particle and a proton are given as $$4 \mathrm{~K}, 2 \mathrm{~K}$$ and $$\mathrm{K}$$ respectively. The de-Broglie wavelength associated with electron $$(\lambda \mathrm{e}), \alpha$$-particle $$((\lambda \alpha)$$ and the proton $$(\lambda p)$$ are as follows:

The threshold frequency of a metal is $$f_{0}$$. When the light of frequency $$2 f_{0}$$ is incident on the metal plate, the maximum velocity of photoelectrons is $$v_{1}$$. When the frequency of incident radiation is increased to $$5 \mathrm{f}_{0}$$, the maximum velocity of photoelectrons emitted is $$v_{2}$$. The ratio of $$v_{1}$$ to $$v_{2}$$ is :