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Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)

Short Answer

Expert verified
The energy of a photon with a wavelength of \(6000 \mathrm{\AA}\) is approximately \(2.07 \mathrm{eV}\), which is closest to option (B), \(2.06 \mathrm{eV}\).

Step by step solution

01

Write down the formula for the energy of a photon

The formula for the energy of a photon is given by: \(E = \dfrac{hc}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (approximated to \(6.63 \times 10^{-34} \mathrm{J·s}\)), \(c\) is the speed of light (approximated to \(3 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon.
02

Convert the given wavelength from angstroms to meters

The wavelength is given as \(6000 \mathrm{\AA}\) (angstroms). To convert this to meters, we can use the conversion factor \(1 \mathrm{\AA} = 10^{-10} \mathrm{m}\). \(\lambda = 6000 \mathrm{\AA} \times \dfrac{10^{-10} \mathrm{m}}{1 \mathrm{\AA}} = 6000 \times 10^{-10} \mathrm{m}\)
03

Calculate the energy of the photon in joules

Using the formula from Step 1 and the wavelength in meters, we can calculate the energy of the photon in joules: \(E = \dfrac{(6.63 × 10^{-34} \mathrm{J·s}) \times (3 × 10^8 \mathrm{m/s})}{(6000 × 10^{-10} \mathrm{m})}\) \(E = \dfrac{1.989 × 10^{-25} \mathrm{J·m}}{6 × 10^{-7} \mathrm{m}}\) \(E = 3.315 × 10^{-19} \mathrm{J}\)
04

Convert energy from joules to electron volts

To convert the energy from joules to electron volts, we can use the conversion factor \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\): \(E_\mathrm{eV} = \dfrac{3.315 \times 10^{-19} \mathrm{J}}{1.602 \times 10^{-19} \mathrm{J/eV}}\) \(E_\mathrm{eV} \approx 2.07 \mathrm{eV}\)
05

Determine the correct answer from the options given

Based on our calculations, the correct answer is approximately \(2.07 \mathrm{eV}\) which is closest to option (B). So the correct answer is: (B) \(2.06 \mathrm{eV}\)

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Most popular questions from this chapter

\(2 \mathrm{nW}\) light of wave length \(4400 \AA\) is incident on photo sensitive surface of \(\mathrm{Cs}\). If quantum efficiency is \(0.5 \%\), what will be the value of photoelectric current? (A) \(1.56 \times 10^{-6} \mu \mathrm{A}\) (B) \(2.56 \times 10^{-6} \mu \mathrm{A}\) (C) \(4.56 \times 10^{-6} \mu \mathrm{A}\) (D) \(3.56 \times 10^{-6} \mu \mathrm{A}\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Energy of photon of wavelength \(\lambda\) is (P) \((\mathrm{E} / \mathrm{p})\) (II) The de Broglie wavelength associated (Q) \(\left(\mathrm{hf} / \mathrm{c}^{2}\right)\) with particle of momentum \(\mathrm{P}\) is (II) Mass of photon in motion is (R) (hc \(/ \lambda\) ) (IV) The velocity of photon of energy (S) \((\mathrm{h} / \mathrm{p})\) \(\mathrm{E}\) and momentum \(\mathrm{P}\) is (A) I - P, II - Q. III - R, IV - S (B) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$ (C) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{P}_{3} \mathrm{IV}-\mathrm{Q}$ (D) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$

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) \(1.534\)

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) \(1.84 \mu \mathrm{A}\)

Output power of He-Ne LASER of low energy is \(1.00 \mathrm{~mW}\). Wavelength of the light is \(632.8 \mathrm{~nm}\). What will be the number of photons emitted per second from this LASER? (A) \(8.31 \times 10^{15} \mathrm{~s}^{-1}\) (B) \(5.38 \times 10^{15} \mathrm{~s}^{-1}\) (C) \(1.83 \times 10^{15} \mathrm{~s}^{-1}\) (D) \(3.18 \times 10^{15} \mathrm{~s}^{-1}\)

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