The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $$6630$$ Å is $$0.42$$ V. If the threshold frequency is $$x \times 10^{13}$$ s, where $$x$$ is (nearest integer): (Given, speed light $$= 3 \times 10^8$$ m s$$^{-1}$$, Planck's constant $$= 6.63 \times 10^{-34}$$ J s)

We need to find the threshold frequency for the photoelectric effect.

Wavelength of incident light, $$\lambda = 6630$$ Å $$= 6630 \times 10^{-10}$$ m

Stopping potential, $$V_0 = 0.42$$ V

Speed of light, $$c = 3 \times 10^8$$ m/s

Planck's constant, $$h = 6.63 \times 10^{-34}$$ J s

$$h\nu = h\nu_0 + eV_0$$

$$h\nu_0 = h\nu - eV_0$$

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6630 \times 10^{-10}} = \frac{3 \times 10^8}{6.63 \times 10^{-7}}$$

$$= 4.525 \times 10^{14}$$ Hz

$$h\nu = 6.63 \times 10^{-34} \times 4.525 \times 10^{14} = 3.0 \times 10^{-19}$$ J

$$eV_0 = 1.6 \times 10^{-19} \times 0.42 = 0.672 \times 10^{-19}$$ J

$$h\nu_0 = 3.0 \times 10^{-19} - 0.672 \times 10^{-19} = 2.328 \times 10^{-19}$$ J

$$\nu_0 = \frac{2.328 \times 10^{-19}}{6.63 \times 10^{-34}} = 3.51 \times 10^{14}$$ Hz

$$= 35.1 \times 10^{13}$$ s$$^{-1}$$

Rounding to the nearest integer: $$x = 35$$.

The value of $$x$$ is 35.