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Chapter 17: Problem 2355

\(11 \times 10^{11}\) Photons are incident on a surface in \(10 \mathrm{~s}\). These photons correspond to a wavelength of \(10 \AA\). If the surface area of the given surface is \(0.01 \mathrm{~m}^{2}\), the intensity of given radiations is \(\ldots \ldots\) $\left\\{\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(21.86 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (B) \(2.186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (C) \(218.6 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) (D) \(2186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\)

Short Answer

Expert verified
The intensity of the given radiations is \(2.186 \times 10^{-3} ~\mathrm{W.m^{-2}}\).

Step by step solution

01

Calculate the energy of a single photon

According to the Planck's equation, the energy (E) of a photon can be determined using the formula: \(E = \dfrac{hc}{\lambda}\) where h = Planck's constant = \(6.625 \times 10^{-34}~J.s\) c = speed of light = \(3 \times 10^8~m.s^{−1}\) λ = wavelength of photon = \(10 \mathrm{~\mathring{A}} = 10^{-10}~m\) Now, substitute these values in the formula and calculate the energy of a single photon: \(E = \dfrac{(6.625 \times 10^{-34}) (3 \times 10^{8})}{10^{-10}}\)
02

Calculate the total energy of all the photons

The total energy can be found by multiplying the energy of a single photon by the total number of photons: Total photonic energy \(=\) Energy of a single photon \(×\) number of photons \(= E \times 11 \times 10^{11}\) Now substitute the expression for E we found in step 1: Total photonic energy \(= \dfrac{(6.625 \times 10^{-34}) (3 \times 10^{8})(11 \times 10^{11})}{10^{-10}}\)
03

Calculate the intensity

Intensity (I) is defined as the total energy incident on an area per unit time. In this case, we are given the area of the surface and the time duration. We can calculate the intensity using the formula: \(I = \dfrac{\text{Total photonic energy}}{\text{Area} \times \text{Time}}\) Substitute the values given in the problem and the expression for total photonic energy derived in step 2: \(I = \dfrac{\dfrac{(6.625 \times 10^{-34}) (3 \times 10^{8})(11 \times 10^{11})}{10^{-10}}}{(0.01)(10)}\)
04

Solve the expression to find the intensity

Now, compute the intensity using the expression above: \(I = \dfrac{(6.625 \times 10^{-34}) (3 \times 10^{8})(11 \times 10^{11})}{(10^{-10})(0.01)(10)} = 2.186 \times 10^{-3} ~\mathrm{W.m^{-2}}\) Upon solving the expression, we find that the intensity of the given radiations is \(2.186 \times 10^{-3} ~\mathrm{W.m^{-2}}\), which corresponds to option (B).

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Most popular questions from this chapter

Photoelectric effect on surface is found for frequencies $5.5 \times 10^{8} \mathrm{MHz}\( and \)4.5 \times 10^{8} \mathrm{MHz}$ If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

Frequency of incident light on body is \(\mathrm{f}\). Threshold frequency of body is \(f_{0}\). Maximum velocity of electron \(=\ldots \ldots \ldots\).. where \(m\) is mass of electron. (A) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]^{(1 / 2)}$ (B) $\left[\left\\{2 \mathrm{~h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\right\\} / \mathrm{m}\right]$ (C) \([2 \mathrm{hf} / \mathrm{m}]^{(1 / 2)}\) (D) \(\mathrm{h}\left(\mathrm{f}-\mathrm{f}_{0}\right)\)

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

An electron enters perpendicularly into uniform magnetic field having magnitude \(0.5 \times 10^{-5} \mathrm{~T}\). If it moves on a circular path of radius \(2 \mathrm{~mm}\), its de - Broglie wavelength is $\ldots \ldots . . . \AA$ (A) 3410 (B) 4140 (C) 2070 (D) 2785

The cathode of a photoelectric cell is changed such that the work function changes from \(\mathrm{W}_{1}\) to \(\mathrm{W}_{2}\left(\mathrm{~W}_{2}>\mathrm{W}_{1}\right)\). If the currents before and after change are \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), all other conditions remaining unchanged, then assuming $\mathrm{hf}>\mathrm{W}_{2} \ldots \ldots$ (A) \(\mathrm{I}_{1}=\mathrm{I}_{2}\) (B) \(I_{1}\mathrm{I}_{2}\) (D) \(\mathrm{I}_{1}<\mathrm{I}_{2}<2 \mathrm{I}_{1}\)
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