In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $$\phi = 2.5$$ eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential.
($$h = 6.63 \times 10^{-34}$$ J s, $$c = 3 \times 10^8$$ m s$$^{-1}$$)

We can calculate the change in stopping potential by applying Einstein’s photoelectric equation to both wavelengths.

1. Einstein’s Photoelectric Equation

The maximum kinetic energy ($$K_{\text{max}}$$) of an emitted electron is the difference between the incident photon energy ($$h\nu$$) and the work function ($$\phi$$) of the metal:

$$K_{\text{max}} = h\nu - \phi$$

Since the stopping potential ($$V_0$$) is the potential required to stop the most energetic electrons, we have:

$$eV_0 = K_{\text{max}}$$

Combining these, and using $$\nu = \frac{c}{\lambda}$$:

$$eV_0 = \frac{hc}{\lambda} - \phi$$

2. Numerical Constants

To simplify the calculation, we use the value of the product $$hc$$ in electron-volt nanometers ($$\text{eV}\cdot\text{nm}$$):

$$\frac{hc}{e} \approx 1240\ \text{eV}\cdot\text{nm}$$

Thus, the energy of a photon ($$E_p$$) in eV is:

$$E_p = \frac{1240}{\lambda \text{ (nm)}}$$

3. Step-by-Step Calculation

For the first wavelength ($$\lambda_1 = 280$$ nm):

• Photon Energy ($$E_1$$): $$\frac{1240}{280} \approx 4.43\ \text{eV}$$
• Max Kinetic Energy ($$K_{\text{max},1}$$): $$4.43\ \text{eV} - 2.5\ \text{eV} = 1.93\ \text{eV}$$
• Stopping Potential ($$V_{0,1}$$): $$1.93\ \text{V}$$
• Photon Energy ($$E_2$$): $$\frac{1240}{400} = 3.10\ \text{eV}$$
• Max Kinetic Energy ($$K_{\text{max},2}$$): $$3.10\ \text{eV} - 2.5\ \text{eV} = 0.60\ \text{eV}$$
• Stopping Potential ($$V_{0,2}$$): $$0.60\ \text{V}$$

For the second wavelength ($$\lambda_2 = 400$$ nm):

4. Change in Stopping Potential ($$\Delta V$$)

The decrease in stopping potential when switching from $$280$$ nm to $$400$$ nm is:

$$\Delta V = V_{0,1} - V_{0,2}$$

$$\Delta V = 1.93\ \text{V} - 0.60\ \text{V} = 1.33\ \text{V}$$

Conclusion:

Rounding to two significant figures, the change is:

$$\boxed{\Delta V \approx 1.3\ \text{V}}$$

This corresponds to Option C.