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Chapter 17: Problem 2336

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

Short Answer

Expert verified
The short answer is: The threshold wavelength of the metal is \(5439 \AA\) (Option C).

Step by step solution

01

Recall the photoelectric effect equation and Planck's equation

For the photoelectric effect, we know that the maximum kinetic energy of ejected electrons is given by: \(K_{max} = hf - hf_0\) where \(K_{max}\): maximum kinetic energy of ejected electrons \(h\): Planck's constant \(f\): frequency of incident radiation \(f_0\): threshold frequency The maximum kinetic energy of ejected electrons can also be expressed as: \(K_{max} = eV\) where \(e\): elementary charge \(V\): stopping potential Planck's equation relates the energy of a photon to its frequency: \(E = hf\) with \(E = \dfrac{hc}{\lambda}\) where \(\lambda\): wavelength of incident radiation \(c\): speed of light
02

Calculate the energy of the incident radiation

We are given the wavelength of the incident radiation, \(\lambda = 3000 \AA\), and we're supposed to find the energy of this radiation using the Planck's equation: \(E = \dfrac{hc}{\lambda}\) Plugging in the values of Planck's constant and speed of light, \(E = \dfrac{(6.6 \times 10^{-34} J.s)(3 \times 10^8 m/s)}{3000 \times 10^{-10} m}\) Calculate the value of energy: \(E = 6.6 \times 10^{-19} J\)
03

Calculate the maximum kinetic energy of ejected electrons

We're given the stopping potential, \(V = 1.85 V\), so we can calculate the maximum kinetic energy of ejected electrons using the relationship \(K_{max} = eV\). Since the elementary charge, \(e = 1.6 \times 10^{-19} C\), \(K_{max} = (1.6 \times 10^{-19} C)(1.85 V)\) Calculate the value of the maximum kinetic energy: \(K_{max} = 2.96 \times 10^{-19} J\)
04

Calculate the energy difference between the incident radiation and the ejected electrons

Use the photoelectric effect equation to find the energy difference: \(hf - hf_0 = K_{max}\) Since we know the incident energy from Step 2: \(hf - hf_0 = 6.6 \times 10^{-19} J - hf_0 = 2.96 \times 10^{-19} J\) Therefore, the threshold energy is \(hf_0 = 3.64 \times 10^{-19} J\)
05

Calculate the threshold wavelength

Now we have the threshold energy, and we can use the Planck's equation to calculate the threshold wavelength: \(\lambda_0 = \dfrac{hc}{hf_0}\) Plugging in the values for Planck's constant, speed of light, and the threshold energy: \(\lambda_0 = \dfrac{(6.6 \times 10^{-34} J.s)(3 \times 10^8 m/s)}{3.64 \times 10^{-19} J}\) Calculate the value of the threshold wavelength: \(\lambda_0 = 5.439 \times 10^{-7} m\)
06

Convert the threshold wavelength to angstrom and choose the correct option

Finally, convert the threshold wavelength from meters to angstrom: \(\lambda_0 = 5.439 \times 10^{-7} m \times \dfrac{10^{10} \AA}{1 m} = 5439 \AA\) So, the correct option is (C) \(5439 \AA\).

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Most popular questions from this chapter

Power produced by a star is \(4 \times 10^{28} \mathrm{~W}\). If the average wavelength of the emitted radiations is considered to be \(4500 \AA\) the number of photons emitted in \(1 \mathrm{~s}\) is \(\ldots \ldots\) (A) \(1 \times 10^{45}\) (B) \(9 \times 10^{46}\) (C) \(8 \times 10^{45}\) (D) \(12 \times 10^{46}\)

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

Energy of a particle having de-Broglie wavelength \(0.004 \AA\) is... (A) \(1280 \mathrm{eV}\) (B) \(1200 \mathrm{eV}\) (C) \(1200 \mathrm{MeV}\) (D) \(1200 \mathrm{GeV}\)

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) \(69.3 \AA\)

At \(10^{\circ} \mathrm{C}\) temperature, de-Broglie wave length of atom is $0.4 \AA\(.If temperature of atom is increased by \)30^{\circ} \mathrm{C}$, what will be change in de-Broglie wavelength of atom? (A) decreases \(10^{-2} \AA\) (B) decreases \(2 \times 10^{-2} \AA\) (C) increases \(10^{-2} \AA\) (D) increases \(2 \times 10^{-2} \AA\)
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