Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $$x : 3$$. The value of $$x$$ is

Let the work function of the metal be $$\phi$$. The two streams of photons have energies $$E_1 = 2\phi$$ and $$E_2 = 10\phi$$.

By the photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is $$KE_{max} = E - \phi$$.

For the first stream: $$\frac{1}{2}mv_1^2 = 2\phi - \phi = \phi$$.

For the second stream: $$\frac{1}{2}mv_2^2 = 10\phi - \phi = 9\phi$$.

Taking the ratio: $$\frac{v_1^2}{v_2^2} = \frac{\phi}{9\phi} = \frac{1}{9}$$.

Therefore, $$\frac{v_1}{v_2} = \frac{1}{3}$$, which means $$v_1 : v_2 = 1 : 3$$.

Comparing with the given ratio $$x : 3$$, we get $$x = 1$$.