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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

If \(\propto\) -particle and proton are accelerated through the same potential difference, then the ratio of de Brogile wavelength of \(\propto\) -particle and proton is \(\ldots \ldots \ldots\) (A) \((1 / \sqrt{2})\) (B) \(\sqrt{2}\) (C) \(\\{1 /(2 \sqrt{2})\\}\) (D) \(2 \sqrt{2}\)

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) \(10^{6} \mathrm{~ms}^{-1}\)

If ratio of threshold frequencies of two metals is \(1: 3\), ratio of their work functions is \(\ldots \ldots\) (A) \(1: 3\) (B) \(3: 1\) (C) \(4: 16\) (D) \(16: 4\)

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is \(\ldots \ldots\) (A) \((\overline{\mathrm{h}} / \lambda)\) (B) \((2 \overline{\mathrm{h}} / 3 \lambda)\) (C) \((\lambda \lambda) \overline{\mathrm{h}}\) (D) \((3 N 2 \lambda) \overline{\mathrm{h}}\)
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