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Chapter 17: Problem 2370

Work function of metal is \(4.2 \mathrm{eV}\) If ultraviolet radiation (photon) having energy \(6.2 \mathrm{eV}\), stopping potential will be........ (A) \(2 \mathrm{eV}\) (B) \(2 \mathrm{~V}\) (C) 0 (d) \(10.4 \mathrm{~V}\)

Short Answer

Expert verified
The stopping potential for a metal with a work function of \(4.2 \mathrm{eV}\) when ultraviolet radiation of \(6.2 \mathrm{eV}\) is incident on it is (B) \(2 \mathrm{~V}\).

Step by step solution

01

List the given information

The given information is: Work function (W) = \(4.2\mathrm{eV}\) Photon energy (E) = \(6.2\mathrm{eV}\)
02

Calculate the energy difference

The energy difference (E - W) is the energy provided by the photon that is not used to overcome the work function. This energy is given by: E - W = \(6.2\mathrm{eV} - 4.2\mathrm{eV} = 2\mathrm{eV}\)
03

Convert eV to Joules

Since we will need to convert the energy difference into voltage, we first need to convert the energy from electron volts (eV) to Joules (J). The conversion factor is: \(1\mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\) Therefore, the energy difference in Joules is: \(2\mathrm{eV} \times 1.6 \times 10^{-19}\frac{\mathrm{J}}{\mathrm{eV}} = 3.2 \times 10^{-19}\mathrm{J}\)
04

Calculate the stopping potential

Now, we can find the stopping potential using the formula: Stopping potential = (Photon energy - Work function) / Elementary charge Here, the elementary charge (q) is the charge of an electron, which is \(1.6 \times 10^{-19} \mathrm{C}\). Plugging in the values we have: Stopping potential = \(\frac{3.2 \times 10^{-19}\mathrm{J}}{1.6 \times 10^{-19} \mathrm{C}} = 2\mathrm{V}\)
05

Check the answer choices

The stopping potential we calculated is \(2\mathrm{V}\). Checking the answer choices, option (B) matches with our calculated value. Hence, the correct answer is: (B) \(2 \mathrm{~V}\)

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Most popular questions from this chapter

Light of \(4560 \AA 1 \mathrm{~mW}\) is incident on photo-sensitive surface of \(\mathrm{Cs}\) (Cesium). If the quantum efficiency of the surface is \(0.5 \%\) what is the amount of photoelectric current produced? (A) \(1.84 \mathrm{~mA}\) (B) \(4.18 \mu \mathrm{A}\) (C) \(4.18 \mathrm{~mA}\) (D) \(1.84 \mu \mathrm{A}\)

Kinetic energy of proton accelerated under p.d. \(1 \mathrm{~V}\) will be........ (A) \(1840 \mathrm{eV}\) (B) \(13.6 \mathrm{eV}\) (C) \(1 \mathrm{eV}\) (D) \(0.54 \mathrm{eV}\)

Wavelength of an electron having energy \(10 \mathrm{ke} \mathrm{V}\) is $\ldots \ldots . \AA$ (A) \(0.12\) (B) \(1.2\) (C) 12 (D) 120

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)

Work functions for tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength of sodium is \)5460 \AA$, threshold wavelength for tungsten will be \(\ldots \ldots . . \AA\) (A) 528 (B) 10683 (C) 2791 (D) 5893
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