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Chapter 17: Problem 2351

U. V. light of wavelength \(200 \mathrm{~nm}\) is incident on polished surface of Fe. work function of the surface is \(4.5 \mathrm{eV}\). Find maximum speed of phote electrons. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(7.75 \times 10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(875 \times 10^{5}(\mathrm{~m} / \mathrm{s})\) (C) \(8.75 \times 10^{4}(\mathrm{~m} / \mathrm{s})\) (D) \(7.75 \times 10^{5}(\mathrm{~m} / \mathrm{s})\)

Short Answer

Expert verified
The maximum speed of the photoelectrons is \(8.75 \times 10^5 \frac{m}{s}\).

Step by step solution

01

Calculate the energy of incident photons

First, we need to calculate the energy of the incident photons using the wavelength and Planck's constant (h). The formula for the energy of a photon is: \[E = \frac{hc}{\lambda}\] where E is the energy, h is Planck's constant (6.625 × 10⁻³⁴ J.s), c is the speed of light (3 × 10⁸ m/s), and λ is the wavelength (200 nm).
02

Convert wavelength to meters

To use the values in SI units, we need to convert the wavelength from nanometers to meters: \[\lambda = 200 \, nm \times \frac{1 \, m}{10^9 \, nm} = 2\times10^{-7}\,m\]
03

Calculate the energy of incident photons in Joules

Now we can calculate the energy of the photons: \[E = \frac{6.625 \times 10^{-34}\, J.s \times 3 \times 10^8 \, m/s}{2 \times 10^{-7} \, m} = 9.9375 \times 10^{-19} J\]
04

Convert energy from Joules to electron volts

Now we need to convert the energy of the photons from Joules to electron volts (eV) using the conversion factor 1 eV = 1.6 × 10⁻¹⁹ J: \[E = 9.9375 \times 10^{-19} \, J \times \frac{1\, eV}{1.6 \times 10^{-19} \, J} \approx 6.21\,eV\]
05

Calculate the maximum kinetic energy of emitted photoelectrons

Using the photoelectric effect equation, we can calculate the maximum kinetic energy (K_max) of the emitted photoelectrons: \[K_{max} = E - W\] where E is the energy of photons (6.21 eV) and W is the work function (4.5 eV). \[K_{max} = 6.21\, eV - 4.5 \, eV = 1.71 \, eV\]
06

Convert the maximum kinetic energy to Joules

Now we need to convert the maximum kinetic energy of photoelectrons back to Joules using the conversion factor 1 eV = 1.6 × 10⁻¹⁹ J: \[K_{max} = 1.71 \, eV \times 1.6 \times 10^{-19} \, J / eV = 2.736 \times 10^{-19} \, J\]
07

Calculate the maximum speed of photoelectrons

Finally, we can find the maximum speed (v_max) of the photoelectrons using the kinetic energy formula: \[K_{max} = \frac{1}{2} m v_{max}^2 \] where m is the mass of an electron (9.109 × 10⁻³¹ kg), and K_max is the maximum kinetic energy (2.736 × 10⁻¹⁹ J). Rearranging the equation, we get: \[v_{max} = \sqrt{\frac{2 K_{max}}{m}}\] Now plug in the values and calculate the maximum speed: \[v_{max} = \sqrt{\frac{2 \times 2.736 \times 10^{-19}\, J}{9.109 \times 10^{-31} \, kg}} \approx 8.75 \times 10^5 \frac{m}{s}\] Thus, the maximum speed of the photoelectrons is 8.75 × 10⁵ m/s, which corresponds to answer choice (C).

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Most popular questions from this chapter

The mass of a particle is 400 times than that of an electron and charge is double. The particle is accelerated by \(5 \mathrm{~V}\). Initially the particle remained at rest, then its final kinetic energy is \(\ldots \ldots \ldots\) (A) \(5 \mathrm{eV}\) (B) \(10 \mathrm{eV}\) (C) \(100 \mathrm{eV}\) (D) \(2000 \mathrm{eV}\)

Suppose \(\Psi(\mathrm{x}, \mathrm{y}, \mathrm{z})\) represents a particle in three dimensional space, then probability of finding the particle in the unit volume at a given point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is $\ldots \ldots$ (A) inversely proportional to $\Psi^{\prime}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (B) directly proportional \(\Psi^{*}\) (C) directly proportional to \(\mid \Psi \Psi^{*}\) (D) inversely proportional to \(\left|\Psi \Psi^{*}\right|\)

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In an experiment to determine photoelectric characteristics for a metal the intensity of radiation is kept constant. Starting with threshold frequency. Now, frequency of incident radiation is increased. It is observed that $\ldots \ldots \ldots$ (A) the number of photoelectrons increases (B) the energy of photoelectrons decreases (C) the number of photoelectrons decreases (D) the energy of photoelectrons increases.
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