In photoelectric experiment keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased,
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
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