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A photon of energy \(1.75 \mathrm{eV}\) has a wavelength of most nearly (A) \(710 \mathrm{~nm}\) (B) \(700 \mathrm{~nm}\) (C) \(650 \mathrm{~nm}\) (D) \(600 \mathrm{~nm}\) (E) \(550 \mathrm{~nm}\)

Short Answer

Expert verified
Answer: (A) 710 nm

Step by step solution

01

Use Planck's formula

Planck's formula relates the energy of a photon (E) to its wavelength (λ) and frequency (ν): \(E = hν\), where h is Planck's constant \((6.626 \times 10^{-34} \mathrm{J\cdot s})\). We also know the relationship between the wavelength and frequency of a photon is given by \(ν = \dfrac{c}{λ} \), where c is the speed of light \((3 \times 10^8 \mathrm{\,m/s})\). Combining these two equations, we can find the relationship between energy and wavelength: \(E = h\left(\dfrac{c}{λ}\right) \) Now solve for λ: \(λ = \dfrac{hc}{E}\)
02

Plug in the given values and convert units

Given energy E = 1.75 eV, we need to convert eV to Joules (J): \(E = 1.75 \mathrm{eV} \times \dfrac{1.6 \times 10^{-19} \mathrm{J}}{1 \mathrm{eV}} = 2.8 \times 10^{-19} \mathrm{J}\) Now plug in the values of h, c, E into the equation: \(λ = \dfrac{(6.626 \times 10^{-34}\mathrm{J\cdot s})(3 \times 10^8 \mathrm{m/s})}{2.8 \times 10^{-19}\mathrm{J}}\)
03

Calculate the wavelength and convert to nanometers

Calculate the wavelength in meters: \(λ = \dfrac{(6.626 \times 10^{-34}\mathrm{J\cdot s})(3 \times 10^8 \mathrm{m/s})}{2.8 \times 10^{-19}\mathrm{J}} = 7.115 \times 10^{-7}\mathrm{m}\) Convert this wavelength to nanometers: \(λ = 7.115 \times 10^{-7}\mathrm{m} \times \dfrac{10^9 \mathrm{nm}}{1 \mathrm{m}} = 711.5 \mathrm{nm}\)
04

Match the result to the given options

Compare the calculated wavelength (711.5 nm) to the options: (A) 710 nm (B) 700 nm (C) 650 nm (D) 600 nm (E) 550 nm The closest value to the calculated wavelength is \(710 \mathrm{nm}\). Therefore, the correct answer is (A).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's Constant and its Role in Photon Energy
Planck's constant is a fundamental quantity in physics that acts as a bridge between the quantum and classical worlds.
It is represented by the symbol "h" and has the value of approximately \(6.626 \times 10^{-34} \mathrm{J \cdot s}\).
In quantum mechanics, it plays a crucial role in the description of the size of quanta.

When dealing with photons, Planck's constant appears in the formula \(E = hu\), where \(E\) is the energy of a photon and \(u\) is its frequency.
This directly shows how energy and frequency are related for photons. Without Planck's constant, we could not quantify the small packets of energy that photons carry.

Understanding its role helps us appreciate the discrete nature of energy at a microscopic scale, making it essential in explaining many physical phenomena such as black body radiation and the photoelectric effect.
Linking Energy and Wavelength of Photons
In the realm of photon energy, the relationship between energy \(E\) and wavelength \(\lambda\) is vital to understand.
Most commonly, this relationship is given by the equation \(E = \dfrac{hc}{\lambda}\).
Here, \(h\) represents Planck's constant and \(c\) is the speed of light, approximately \(3 \times 10^8 \mathrm{m/s}\).

This equation highlights an inverse relationship:
  • As wavelength increases, energy decreases.
  • As wavelength decreases, energy increases.
Understanding this connection allows us to deduce the properties of light in various parts of the electromagnetic spectrum, from radio waves (long wavelength, low energy) to gamma rays (short wavelength, high energy).
With this fundamental relation, we can calculate the energy associated with any given wavelength of light, critical for fields such as spectroscopy and quantum mechanics.
Converting Units in Physics Calculations
Unit conversion is a vital skill in physics to ensure accuracy in calculations.
In our problem, we encountered energy given in electron volts (\(\mathrm{eV}\)), which we needed to convert to joules ( \(\mathrm{J}\)).
The conversion factor between these units is \(1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\).

Here's why unit conversion matters:
  • It ensures the consistency of units across different measurements.
  • Correct unit usage allows for direct application in standard equations used in physics.
For wavelength, converting from meters to nanometers involves using the conversion factor \(1 \mathrm{m} = 10^9 \mathrm{nm}\).
Efficient unit conversion is crucial to problem solving, as minor inaccuracies can lead to large errors in results.
Always check units are matching in your calculations to maintain precision and validity.

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Most popular questions from this chapter

. Who of the following did not pioneer the study of radioactivity? (A) Marie Curie (B) Pierre Curie (C) Albert Einstein (D) Felix Abel Niepce de Saint Victor (E) Henri Becquerel

Two isotopes of an element ordinarily (A) have the same atomic number. (B) have the same atomic mass. (C) contain the same number of nucleons in the nucleus. (D) contain the same number of electrons. (E) (A) and (D)

The half-life of the hydrogen isotope tritium is about 12 years. After a certain amount of time a fraction \(31 / 32\) of the atoms in the original sample has decayed. The time is most nearly equal to (A) 12 years (B) 24 years (C) 36 years (D) 48 years (E) 60 years

Measurements of the spectrum of a certain atom show that one sequence of spectral lines has the identical pattern of the Balmer series except that all the frequencies are exactly four times as high. The most likely explanation for the observation is that (A) one is observing a series of lines in hydrogen that have the ground state \(n=1\) instead of \(n=2\). (B) the atom is helium, with atomic number two. (C) the atom is beryllium with atomic number four. (D) the atom is deuterium with atomic mass two. (E) the atom is hydrogen but excited to energy levels that are four times higher than in the Balmer series.

Which statements are true about Max Planck's quantum postulate? (More than one answer permitted.) (A) The minimum energy of radiation emitted by a system can come only in discrete, or discontinuous, values. (B) The energy radiated by a system can come only in integral multiples of Planck's constant. (C) The energy radiated by a system can come only in integral multiples of Planck's constant multiplied by the frequency of the emitted light. (D) The energy levels of a system are equally spaced by intervals of Planck's constant multiplied by the frequency of light. (E) The energy radiated from a system must be in the form of photons.

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