In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: $$\left(\frac{hc}{e} = 1240 \text{ nm} \cdot V\right)$$

The photoelectric equation, first stated by Einstein, relates the stopping potential $$V_0$$ to the frequency $$\nu$$ (or wavelength $$\lambda$$) of the incident light as

$$eV_0 \;=\; h\nu \;-\;\phi_0$$

where $$e$$ is the electronic charge, $$h$$ is Planck’s constant, and $$\phi_0$$ is the work function of the metal. Using the relation between frequency and wavelength $$\nu = \dfrac{c}{\lambda}$$, this can be rewritten in terms of wavelength:

$$eV_0 \;=\; \dfrac{hc}{\lambda} \;-\;\phi_0$$

We perform the experiment at two different wavelengths, $$\lambda_1 = 300 \text{ nm}$$ and $$\lambda_2 = 400 \text{ nm}$$, and obtain two corresponding stopping potentials $$V_1$$ and $$V_2$$. Writing the equation for each wavelength, we have

$$eV_1 \;=\; \dfrac{hc}{\lambda_1} \;-\;\phi_0$$

$$eV_2 \;=\; \dfrac{hc}{\lambda_2} \;-\;\phi_0$$

To eliminate the unknown work function $$\phi_0$$, we subtract the second equation from the first:

$$eV_1 \;-\; eV_2 \;=\;\dfrac{hc}{\lambda_1} \;-\;\dfrac{hc}{\lambda_2}$$

Taking $$e$$ common on the left side, this becomes

$$e\,(V_1 - V_2) \;=\; \dfrac{hc}{\lambda_1} \;-\; \dfrac{hc}{\lambda_2}$$

Dividing by $$e$$ gives the change (decrease) in stopping potential:

$$\Delta V \;=\; V_1 - V_2 \;=\; \dfrac{hc}{e}\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$$

We are supplied with the convenient constant $$\dfrac{hc}{e} = 1240 \text{ nm}\cdot\text{V}$$. Substituting the given wavelengths, we obtain

$$\Delta V \;=\; 1240 \,\text{nm}\cdot\text{V}\;\Bigl(\dfrac{1}{300 \,\text{nm}} - \dfrac{1}{400 \,\text{nm}}\Bigr)$$

Now we evaluate the term in parentheses carefully, keeping every algebraic step explicit:

$$\dfrac{1}{300 \,\text{nm}} \;=\; 0.003333\;\text{nm}^{-1}$$

$$\dfrac{1}{400 \,\text{nm}} \;=\; 0.002500\;\text{nm}^{-1}$$

Hence,

$$\dfrac{1}{300} - \dfrac{1}{400} \;=\; 0.003333 - 0.002500 \;=\; 0.000833\;\text{nm}^{-1}$$

Multiplying by the constant 1240 nm·V, we get

$$\Delta V \;=\; 1240 \times 0.000833 \;\text{V}$$

Carrying out the multiplication step by step,

$$1240 \times 0.000833 \;=\; 1.033\;\text{V (approximately)}$$

The decrease in the stopping potential is therefore about $$1.0 \text{ V}$$ when expressed to one significant figure, matching the choice closest to our calculated value.

Hence, the correct answer is Option C.