de-Broglie wavelength associated with an electron accelerated through a potential difference '$$\mathrm{V}$$' is '$$\lambda$$'. When the accelerating potential is increased to '$$4 \mathrm{~V}$$', de-Broglie wavelength.

In photoelectric effect, the photo current

According to de-Broglie hypothesis if an electron of mass '$$m$$' is accelerated by potential difference '$$V$$', the associated wavelength is '$$\lambda$$'. When a proton of mass '$$\mathrm{M}$$' is accelerated through potential difference $$9 \mathrm{~V}$$, then the wavelength associated with it is

When wavelength of incident radiation on the metal surface is reduced from '$$\lambda_1$$' to '$$\lambda_2$$', the kinetic energy of emitted photoelectrons is tripled. The work function of metal [$$\mathrm{h}=$$ Plank's constant, $$\mathrm{c}=$$ velocity of light]