Given, Initial wavelength, $\lambda_1 = 250 \times 10^{-9} \text{ m}$ Initial stopping potential, $V_{01} = 4 \text{ V}$ Final wavelength, $\lambda_2 = 300 \times 10^{-9} \text{ m}$ Planck's constant, $h = 6.626 \times 10^{-34} \text{ J}\,\text{s}$ Speed of light, $c = 3 \times 10^8 \text{ m/s}$ Charge of an electron, $e = 1.602 \times 10^{-19} \text{ C}$ According to Einstein's photoelectric equation: $$ \begin{aligned} \frac{hc}{\lambda_1} &= W + eV_{01} \quad \dots (i) \\ \frac{hc}{\lambda_2} &= W + eV_{02} \quad \dots (ii) \end{aligned} $$ Subtracting equation (ii) from equation (i): $$ \begin{aligned} \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} &= e(V_{01} - V_{02}) \\ hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) &= e(V_{01} - V_{02}) \end{aligned} $$ Substituting the values: $$ \begin{aligned} (6.626 \times 10^{-34})(3 \times 10^8) \left( \frac{1}{250 \times 10^{-9}} - \frac{1}{300 \times 10^{-9}} \right) &= (1.602 \times 10^{-19})(4 - V_{02}) \\ 1.9878 \times 10^{-25} \left( 4 \times 10^6 - 3.3333 \times 10^6 \right) &= (1.602 \times 10^{-19})(4 - V_{02}) \\ 1.9878 \times 10^{-25} (6.6667 \times 10^5) &= (1.602 \times 10^{-19})(4 - V_{02}) \\ 1.3252 \times 10^{-19} &= (1.602 \times 10^{-19})(4 - V_{02}) \\ 4 - V_{02} &= \frac{1.3252 \times 10^{-19}}{1.602 \times 10^{-19}} \\ 4 - V_{02} &= 0.8272 \\ V_{02} &= 4 - 0.8272 \\ V_{02} &= 3.1728 \text{ V} \end{aligned} $$ Thus, the new stopping potential is $3.1728 \text{ V}$.