1. **State the problem:** We are given the pressure function $$P(n) = P_0 e^{-\frac{\lambda n}{h}}$$ where $$P_0 = 1 \times 10^5$$ Pascals, $$\lambda = 6.2 \times 10^4$$, and $$h = 1250$$ meters. We need to find the rate of change of pressure with respect to altitude $$n$$, i.e., $$\frac{dP}{dn}$$, at $$h = 1250$$ m. 2. **Recall the formula:** The derivative of an exponential function $$f(n) = Ae^{kn}$$ is $$f'(n) = Ake^{kn}$$. 3. **Apply the derivative:** Here, $$P(n) = P_0 e^{-\frac{\lambda n}{h}}$$, so the exponent is $$k = -\frac{\lambda}{h}$$. 4. **Calculate the derivative:** $$\frac{dP}{dn} = P_0 \cdot \left(-\frac{\lambda}{h}\right) e^{-\frac{\lambda n}{h}} = -\frac{\lambda}{h} P_0 e^{-\frac{\lambda n}{h}}$$ 5. **Substitute the given values:** $$P_0 = 1 \times 10^5$$ $$\lambda = 6.2 \times 10^4$$ $$h = 1250$$ 6. **Final expression for the rate of change:** $$\frac{dP}{dn} = -\frac{6.2 \times 10^4}{1250} \times 1 \times 10^5 e^{-\frac{6.2 \times 10^4}{1250} n}$$ 7. **Simplify the constant coefficient:** $$\frac{6.2 \times 10^4}{1250} = \frac{62000}{1250} = 49.6$$ 8. **Therefore:** $$\frac{dP}{dn} = -49.6 \times 10^5 e^{-49.6 n} = -4.96 \times 10^6 e^{-49.6 n}$$ **Answer:** The rate of change of pressure with respect to altitude $$n$$ when $$h=1250$$ m is $$\boxed{\frac{dP}{dn} = -4.96 \times 10^6 e^{-49.6 n} \text{ Pascals per meter}}$$