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 begin with Einstein’s photoelectric equation, which relates the maximum kinetic energy of the emitted electrons to the energy of the incident photons and the work function of the metal:

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

The stopping potential $$V_0$$ is defined by the equality

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

Combining these two relations gives

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

or, in terms of wavelength $$\lambda$$ (because $$\nu = \dfrac{c}{\lambda}$$),

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

Hence, for any wavelength, the stopping potential is

$$V_0 \;=\; \dfrac{1}{e}\Bigl(\dfrac{hc}{\lambda} \;-\;\phi\Bigr)$$

To handle the numerical work most cleanly we first evaluate the factor $$\dfrac{hc}{e}$$ once and for all. Using the given constants

$$h = 6.63\times10^{-34}\ \text{J·s}, \qquad c = 3.0\times10^{8}\ \text{m·s}^{-1}, \qquad e = 1.6\times10^{-19}\ \text{C},$$

we have

$$\dfrac{hc}{e} \;=\; \dfrac{6.63\times10^{-34}\times3.0\times10^{8}}{1.6\times10^{-19}} \;=\; 1.24\times10^{-6}\ \text{eV·m}.$$

Because $$1\ \text{m}=10^{9}\ \text{nm}$$, we usually quote this product as

$$\dfrac{hc}{e} \;=\; 1240\ \text{eV·nm}.$$

Thus the photon energy in electron-volts can be written compactly as

$$E_{\text{photon}}(\text{eV}) \;=\; \dfrac{1240}{\lambda(\text{nm})}.$$

Now we examine the two different wavelengths given in the problem.

1. First wavelength: $$\lambda_1 = 280\ \text{nm}$$

The photon energy is

$$E_1 \;=\; \dfrac{1240}{280}\ \text{eV} \;=\; 4.4286\ \text{eV}.$$

The work function of the lithium cathode is given as $$\phi = 2.5\ \text{eV}$$, so the maximum kinetic energy of the emitted electrons is

$$K_{\text{max},1} \;=\; E_1 - \phi \;=\; 4.4286\ \text{eV} - 2.5\ \text{eV} \;=\; 1.9286\ \text{eV}.$$

The corresponding stopping potential follows from $$eV_{0,1}=K_{\text{max},1}$$, which in electron-volt units is simply

$$V_{0,1} \;=\; 1.9286\ \text{V}.$$

2. Second wavelength: $$\lambda_2 = 400\ \text{nm}$$

The photon energy is

$$E_2 \;=\; \dfrac{1240}{400}\ \text{eV} \;=\; 3.10\ \text{eV}.$$

The maximum kinetic energy now becomes

$$K_{\text{max},2} \;=\; E_2 - \phi \;=\; 3.10\ \text{eV} - 2.5\ \text{eV} \;=\; 0.60\ \text{eV}.$$

Thus the new stopping potential is

$$V_{0,2} \;=\; 0.60\ \text{V}.$$

Change in stopping potential

The change (decrease) when we switch from 280 nm to 400 nm is

$$\Delta V \;=\; V_{0,1} - V_{0,2} \;=\; 1.9286\ \text{V} - 0.60\ \text{V} \;=\; 1.3286\ \text{V}.$$

Rounding to two significant figures, this is

$$\Delta V \;\approx\; 1.3\ \text{V}.$$

Hence, the correct answer is Option C.