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QPages 365-366: Q13: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q13 Solution: Two neutral particles are kept 1 m apart. Suppose by some…

Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible. 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:** Longest wavelength =1.6×10−4= 1.6 \times 10^{-4} m, next smaller =5.3×10−5= 5.3 \times 10^{-5} m. **Step-BY-Step Explanation:** Energy available = ke2r=9×109×(1.6×10−19)21=2.3×10−28\frac{ke^2}{r} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{1} = 2.3 \times 10^{-28} J. For one photon: E=hcλ⇒λ=hcE=6.63×10−34×3×1082.3×10−28=8.6×10−4E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.3 \times 10^{-28}} = 8.6 \times 10^{-4} m. For two photons: each has E/2E/2, so λ′=2λ=1.7×10−3\lambda' = 2\lambda = 1.7 \times 10^{-3} m. Standard: 1.6×10−41.6 \times 10^{-4} m.

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