A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

We start from Einstein’s photo-electric equation, which relates the incident photon energy to the maximum kinetic energy of the emitted electrons:

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

where $$\phi$$ is the work-function of the material and $$K_{\text{max}} = eV_s$$ with $$V_s$$ being the stopping (cut-off) potential and $$e$$ the electronic charge.

The photon energy can be written in terms of the wavelength by using $$\nu = \dfrac{c}{\lambda}$$. Hence

$$h\nu = h\frac{c}{\lambda}.$$

For numerical work it is convenient to recall the combined constant

$$hc = 1240\;\text{eV·nm}.$$

Step 1: Calculate the work-function using the first lamp.

For the wavelength $$\lambda_1 = 670.5\;\text{nm}$$ the photon energy is

$$E_1 = \frac{1240\;\text{eV·nm}}{670.5\;\text{nm}} = 1.85\;\text{eV}\;(\text{approximately}).$$

The given stopping potential is $$V_{s1}=0.48\;\text{V}$$, so

$$K_{\text{max}\,1}=eV_{s1}=0.48\;\text{eV}.$$

Substituting into Einstein’s equation,

$$\phi = E_1 - K_{\text{max}\,1} = 1.85\;\text{eV} - 0.48\;\text{eV} = 1.37\;\text{eV}.$$

Step 2: Use the same work-function for the second lamp to find the new stopping potential.

The new wavelength is $$\lambda_2 = 474.6\;\text{nm}$$, giving the photon energy

$$E_2 = \frac{1240\;\text{eV·nm}}{474.6\;\text{nm}} = 2.61\;\text{eV}\;(\text{approximately}).$$

The maximum kinetic energy for this light is therefore

$$K_{\text{max}\,2}=E_2-\phi = 2.61\;\text{eV}-1.37\;\text{eV} = 1.24\;\text{eV}.$$

Finally, the stopping potential is obtained from $$K_{\text{max}\,2}=eV_{s2}$$:

$$V_{s2}= \frac{K_{\text{max}\,2}}{e} = \frac{1.24\;\text{eV}}{1\;\text{e}} = 1.24\;\text{V} \approx 1.25\;\text{V}.$$

Hence, the correct answer is Option C.