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Chapter 17: Problem 2409

The work function of a metal is \(1 \mathrm{eV}\). Light of wavelength $3000 \AA$ is incident on this metal surface. The maximum velocity of emitted photoelectron will be \(\ldots \ldots .\) (A) \(10 \mathrm{~ms}^{-1}\) (B) \(10^{3} \mathrm{~ms}^{-1}\) (C) \(10^{4} \mathrm{~ms}^{-1}\) (D) \(10^{6} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The maximum velocity of the emitted photoelectrons is approximately \(10^6 \text{~ms}^{-1}\) (option D) when light of wavelength \(3000 \mathring{A}\) is incident on a metal surface with a work function of \(1 \mathrm{eV}.\)

Step by step solution

01

Identify the given data

The work function of the metal (Φ) is given as 1 eV and the wavelength of the incident light (λ) as 3000 Å.
02

Convert given data into SI units and needed units

First, we need to convert the work function of the metal from electronvolt (eV) to Joules (J) and wavelength from Angstrom (Å) to meter (m). Φ = 1 eV × (1.6 × 10^{-19} J/eV) = 1.6 × 10^{-19} J λ = 3000 Å × (1 × 10^{-10} m/Å) = 3 × 10^{-7} m
03

Calculate the energy of the incident photons using Planck's equation

Planck's equation relates energy (E) with the wavelength (λ) of a photon using the formula: E = \(\frac{hc}{\lambda}\) where h is the Planck constant (6.63 × 10^{-34} J s) and c is the speed of light (3 × 10^8 m/s). E = \(\frac{(6.63 \times 10^{-34} \text{J s}) \times (3 \times 10^{8} \text{m/s})}{3 \times 10^{-7} \text{m}}\) E = 6.63 × 10^{-19} J
04

Calculate the maximum kinetic energy of the emitted photoelectrons

Using the photoelectric effect equation, we can find the maximum kinetic energy (K.E.) of the emitted photoelectrons: K.E. = E - Φ K.E. = 6.63 × 10^{-19} J - 1.6 × 10^{-19} J K.E. = 5.03 × 10^{-19} J
05

Calculate the maximum velocity of the photoelectrons

Using the kinetic energy equation, we can find the maximum velocity (v) of the emitted photoelectrons: K.E. = \(\frac{1}{2}mv^{2}\) where m is the mass of an electron (9.11 × 10^{-31} kg). Rearranging the equation to solve for v, we get: v = \( \sqrt{\frac{2 \times \text{K.E.}}{m}} \) v = \( \sqrt{\frac{2 \times 5.03 \times 10^{-19} \text{J}}{9.11 \times 10^{-31} \text{kg}}} \) v ≈ 10^6 m/s
06

Choose the correct answer from the given options

The maximum velocity of the emitted photoelectrons is approximately 10^6 m/s, which corresponds to option (D). Therefore, the correct answer is: (D) \(10^{6} \text{~ms}^{-1}\)

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