The ratio of wavelengths of proton and deuteron accelerated by potential V_p and V_d is 1 : $$\sqrt2$$. Then the ratio of V_p to V_d will be :

A metal exposed to light of wavelength $$800 \mathrm{~nm}$$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $$500 \mathrm{~nm}$$ is used. The workfunction of the metal is : (Take hc $$=1230 \,\mathrm{eV}-\mathrm{nm}$$ ).

A source of monochromatic light liberates 9 $$\times$$ 10²⁰ photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :

The electric field at a point associated with a light wave is given by

E = 200 [sin (6 $$\times$$ 10¹⁵)t + sin (9 $$\times$$ 10¹⁵)t] Vm^$$-$$1

Given : h = 4.14 $$\times$$ 10^$$-$$15 eVs

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be