A monochromatic light loses 25 % of its speed when it enters a transparent medium from air. Determine the refractive index of the medium

The work function $\phi$ is given as $1.82 \, eV$.
We use the conversion factor $1 \, eV = 1.602 \times 10^{-19} \, J$.

$$\phi = 1.82 \, eV \times \left(1.602 \times 10^{-19} \frac{J}{eV}\right) = 2.91564 \times 10^{-19} \, J$$

The relationship between work function ($\phi$), Planck's constant ($h$), speed of light ($c$), and threshold wavelength ($\lambda_0$) is given by:

$$\phi = \frac{hc}{\lambda_0}$$

Rearranging to solve for $\lambda_0$:

$$\lambda_0 = \frac{hc}{\phi}$$

Given values:
$h = 6.63 \times 10^{-34} \, Js$
$c = 3.0 \times 10^8 \, ms^{-1}$
$\phi = 2.91564 \times 10^{-19} \, J$

$$\lambda_0 = \frac{(6.63 \times 10^{-34} \, Js) \times (3.0 \times 10^8 \, ms^{-1})}{2.91564 \times 10^{-19} \, J}$$ $$\lambda_0 = \frac{19.89 \times 10^{-26} \, Jm}{2.91564 \times 10^{-19} \, J}$$ $$\lambda_0 = 6.8212 \times 10^{-7} \, m$$

Since $1 \, nm = 10^{-9} \, m$:

$$\lambda_0 = 6.8212 \times 10^{-7} \, m \times \frac{1 \, nm}{10^{-9} \, m} = 682.12 \, nm$$

Rounding to the nearest whole number, $\lambda_0 \approx 682 \, nm$.

Comparing this to the given options:
A. 682 nm
B. 783 nm
C. 880 nm
D. 982 nm

The calculated threshold wavelength matches option A.

The final answer is $\boxed{\text{A. 682 nm}}$.