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Chapter 17: Problem 2340

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) \(12 \times 10^{46}\)

Short Answer

Expert verified
The number of photons emitted per second by the star is approximately \(9 \times 10^{46}\).

Step by step solution

01

Convert the given wavelength to meters

The given wavelength is in the unit of Angstroms, but we need to convert it to meters to use the Planck's equation properly. To do this, we will use the conversion factor of \(1\,\text{Angstrom} = 10^{-10}\,\text{m}\). Given wavelength: \(4500 \AA\) Conversion to meters: \(4500 \cdot 10^{-10} m = 4.5 \cdot 10^{-7} m\) The new wavelength in meters is \(4.5 \times 10^{-7}\,\text{m}\).
02

Calculate the energy of a single photon

Now, we will use the Planck's equation to calculate the energy of a single photon: \(E = \dfrac{hc}{\lambda}\) Here, \(E\) is the energy of a photon, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{J\cdot s}\)), \(c\) is the speed of light (\(3 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the emitted radiation. \(E = \frac{(6.626 \times 10^{-34}\mathrm{J\cdot s}) (3 \times 10^8\mathrm{m/s})}{4.5 \times 10^{-7}\mathrm{m}}\) Now, calculate the energy: \(E \approx 4.4193 \times 10^{-19}\,\text{J}\) The energy of a single photon is approximately \(4.4193 \times 10^{-19}\, \text{J}\).
03

Determine the number of photons emitted per second

The next step is to find the number of photons emitted per second by dividing the power produced by the star with the energy of a single photon: Number of photons emitted per second = \(\dfrac{\text{Power produced by star}}{\text{Energy of a single photon}}\) Number of photons emitted per second = \(\dfrac{4 \times 10^{28}\,\text{W}}{4.4193 \times 10^{-19}\,\text{J}}\) Now, perform the calculation: Number of photons emitted per second \(\approx 9.05 \times 10^{46}\)
04

Round the result and compare with the given options

Round the result with significant digits to get the final answer: Number of photons emitted per second = \(\approx 9 \times 10^{46}\) Comparing the result with the given options, we find that the answer is (B) \(9 \times 10^{46}\).

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Most popular questions from this chapter

According to Rayleigh and Jeans the black body radiation in the cavity is system of (A) progressive electromagnetic waves (B) standing electromagnetic waves (C) electromagnetic waves of discrete (D) standing waves in lattice frequencies

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

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) \(4 \mathrm{I}\)

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) \(10.4 \mathrm{~V}\)

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)
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