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Chapter 17: Problem 2420

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.

Short Answer

Expert verified
As the frequency of incident radiation is increased while keeping the intensity constant, the energy of photoelectrons increases, according to Einstein's photoelectric equation \(E_k = h(f - f_0)\). However, the number of photoelectrons remains constant due to the constant intensity. So, the correct answer is (D) the energy of photoelectrons increases.

Step by step solution

01

Understand the Photoelectric Effect

The photoelectric effect is a phenomenon where electrons are emitted from the surface of a metal when it is exposed to light. When light of a frequency greater than the threshold frequency (minimum frequency required to initiate the process) is incident on the metal surface, electrons absorb the energy and are ejected. These ejected electrons are called photoelectrons. Step 2: Recall Einstein's Photoelectric Equation
02

Recall Einstein's Photoelectric Equation

Einstein's photoelectric equation is given by the following formula: \[ E_k = h(f - f_0) \] where \(E_k\) is the kinetic energy of the ejected photoelectron, \(h\) is Planck's constant, \(f\) is the frequency of incident radiation, and \(f_0\) is the threshold frequency. Step 3: Analyze the effect of increasing frequency on energy of photoelectrons
03

Analyze the effect of increasing frequency on energy of photoelectrons

According to Einstein's photoelectric equation, the kinetic energy of the ejected photoelectron is directly proportional to the difference between the incident frequency (\(f\)) and threshold frequency (\(f_0\)). As the frequency of incident radiation is increased, the difference (\(f - f_0\)) also increases. Consequently, the kinetic energy of the photoelectrons will increase. Step 4: Analyze the effect of increasing frequency on the number of photoelectrons
04

Analyze the effect of increasing frequency on the number of photoelectrons

The intensity of radiation is kept constant in the given problem. Intensity is proportional to the number of photons incident on the metal surface. In the photoelectric effect, one incident photon results in the emission of one photoelectron (if the frequency is greater than the threshold frequency). Since the intensity is constant, it implies that the number of photons (and therefore, the number of photoelectrons) remains constant. Step 5: Select the correct options
05

Select the correct options

Based on our analysis in Steps 3 and 4, the following options are correct: (A) the number of photoelectrons increases - Incorrect (B) the energy of photoelectrons decreases - Incorrect (C) the number of photoelectrons decreases - Incorrect (D) the energy of photoelectrons increases - Correct Hence, the correct answer is (D) the energy of photoelectrons increases.

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Most popular questions from this chapter

Work function of \(\mathrm{Zn}\) is \(3.74 \mathrm{eV}\). If the sphere of \(\mathrm{Zn}\) is illuminated by the X-ray of wavelength \(12 \AA\) the maximum potential produced on the sphere is \(\ldots \ldots \ldots\) \(\left(\mathrm{h}-6.625 \times 10^{-34 \mathrm{~J}-\mathrm{s}}\right)\) (A) \(10.314 \mathrm{~V}\) (B) \(103.14 \mathrm{~V}\) (C) \(1031.4 \mathrm{~V}\) (D) \(10314 \mathrm{~V}\)

A star which can be seen with naked eye from Earth has intensity $1.6 \times 10^{-9} \mathrm{Wm}^{-2}\( on Earth. If the corresponding wavelength is \)560 \mathrm{~nm}\(, and the radius of the human eye is \)2.5 \times 10^{-3} \mathrm{~m}\(, the number of photons entering in our in \)1 \mathrm{~s}$ is..... (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\) (B) \(8.85 \times 10^{4} \mathrm{~s}^{-1}\) (C) \(7.85 \times 10^{5} \mathrm{~s}^{-1}\) (D) \(8.85 \times 10^{5} \mathrm{~s}^{-1}\)

Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

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) \(2 \sqrt{2}\)

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)
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