When a metallic surface is illuminated with radiation of wavelength '$$\lambda$$', the stopping potential is '$$\mathrm{V}$$'. If the same surface is illuminated with radiation of wavelength '$$2 \lambda$$', the stopping potential is '$$\left(\frac{\mathrm{v}}{4}\right)$$'. The threshold wavelength for the metallic surface is
- A$$\frac{5}{2} \lambda$$
- B$$3 \lambda$$
- C$$4 \lambda$$
- D$$5 \lambda$$
✓ Correct answer: B
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