When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $$v_1$$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $$v_2$$. If $$v_2 = xv_1$$, the value of $$x$$ will be ______.

We use the photoelectric equation: $$h\nu = h\nu_0 + \frac{1}{2}mv^2$$, where $$\nu_0$$ is the threshold frequency.

When the frequency is $$2\nu_0$$, the maximum velocity is $$v_1$$, so

$$ h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2 $$

$$ \frac{1}{2}mv_1^2 = h\nu_0 \quad \cdots (1) $$

When the frequency is $$5\nu_0$$, the maximum velocity is $$v_2$$, so

$$ h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2 $$

$$ \frac{1}{2}mv_2^2 = 4h\nu_0 \quad \cdots (2) $$

Dividing equation (2) by equation (1) gives

$$ \frac{v_2^2}{v_1^2} = \frac{4h\nu_0}{h\nu_0} = 4 $$ and taking the square root yields $$ \frac{v_2}{v_1} = 2 $$.

Since $$v_2 = xv_1$$, we get x = 2.