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QPages 365-366: Q16: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q16 Solution: The work function of a photoelectric material is 4.0 eV. a)…

The work function of a photoelectric material is 4.0 eV. a) What is the threshold wavelength? b) Find the wavelength of light for which the stopping potential is 2.5 V. Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** (a) λ0=310\lambda_0 = 310 nm (b) λ=190\lambda = 190 nm. **Step-by-Step Explanation:** (a) ϕ=4.0\phi = 4.0 eV. λ0=12404.0=310\lambda_0 = \frac{1240}{4.0} = 310 nm. (b) E=ϕ+eVs=4.0+2.5=6.5E = \phi + eV_s = 4.0 + 2.5 = 6.5 eV. λ=12406.5=190.8\lambda = \frac{1240}{6.5} = 190.8 nm.

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