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Chapter 17: Problem 2371

If intensity of incident light is increased, ........of photo electrons will increase. (A) number (B) Frequency (C) energy (D) wavelength

Short Answer

Expert verified
When the intensity of incident light is increased, the \(number\) of photo electrons will increase, not their frequency, energy, or wavelength. The correct answer is (A) number.

Step by step solution

01

Understanding the Photoelectric effect

The Photoelectric effect is the phenomenon of the emission of electrons from a material, typically a metal, when it is exposed to light. In this effect, the energy of the incident light is transferred to electrons in the material. It's crucial to understand that the energy of each individual photon of light is given by the equation \(E = hf\), where \(h\) is Plank's constant and \(f\) is the frequency of light. This means that the energy of each photoelectron (which is an electron that has absorbed the energy of a light photon) is dependent on the frequency of the incoming light, not its intensity.
02

Effect of increasing the intensity of light

The intensity of light is basically a measure of the number of photons coming from the light source per unit time. Thus, increasing the intensity means increasing the number of photons that are incident on the material per unit time. Each of these photons can potentially interact with an electron in the material and give it enough energy to escape, thereby increasing the number of photoelectrons.
03

Understanding the characteristics of photoelectrons

However, it's important to note that while the number of photoelectrons will increase, the energy, frequency, and wavelength of each individual photoelectron does not change with an increase in light intensity. These characteristics are only changed by altering the frequency of the incoming light, as shown by the equation \(E = hf\).
04

Conclusion

So, when the intensity of incident light is increased, the "number" of photo electrons will increase, not their frequency, energy or wavelength. This means that the correct answer is (A) number.

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Most popular questions from this chapter

Photons of energy \(1 \mathrm{eV}\) and \(2.5 \mathrm{ev}\) successively illuminate a metal, whose work function is \(0.5 \mathrm{eV}\), the ratio of maximum speed of emitted election is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(3: 1\) (D) \(1: 3\)

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)

Energy of photon having wavelength \(\lambda\) is \(2 \mathrm{eV}\). This photon when incident on metal. Maximum velocity of emitted is \(\mathrm{V}\). If \(\lambda\) is decreased \(25 \%\) and maximum velocity is made double, work function of metal is \(\ldots \ldots \ldots . . \mathrm{V}\) (A) \(1.2\) (B) \(1.5\) (C) \(1.6\) (D) \(1.8\)

An electron is accelerated under a potential difference of \(64 \mathrm{~V}\), the de Broglie wave length associated with electron is $=\ldots \ldots \ldots \ldots . \AA$ (Use charge of election $=1.6 \times 10^{-19} \mathrm{C},=9.1 \times 10^{-31} \mathrm{Kg}$, mass of electrum \(\mathrm{h}=6.6623 \times 10^{-43} \mathrm{~J}\).sec \()\) (A) \(4.54\) (B) \(3.53\) (C) \(2.53\) (D) \(1.534\)

An image of sun is formed by a lens of focal length \(30 \mathrm{~cm}\) on the metal surface of a photo-electric cell and a photoelectric current (I) is produced. The lens forming the image is then replaced by another of the same diameter but of focal length of \(15 \mathrm{~cm}\). The photoelectric current in this case is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) 1 (C) \(2 \mathrm{I}\) (D) \(4 \mathrm{I}\)
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