Photons of energy $$10 \mathrm{~eV}$$ are incident on a photosensitive surface of threshold frequency $$2 \times 10^{15} \mathrm{~Hz}$$. The kinetic energy in $$\mathrm{eV}$$ of the photoelectrons emitted is

[Planck's constant $$\mathrm{h}=6.63 \times 10^{34} \mathrm{~Js}$$ ]

The wave number of the last line of the Balmer series in the hydrogen spectrum will be $$\left(\right.$$ Rydberg's cons $$\left.\tan t, R=\frac{10^7}{\mathrm{~m}}\right)$$

Photoemission from metal surface takes place for frequencies '$$v_1$$' and '$$v_2$$' of incident rays $$\left(v_1>v_2\right)$$. Maximum kinetic energy of photoelectrons emitted is in the ratio $$1: \mathrm{K}$$. The threshold frequency of metallic surface is

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)