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Chapter 17: Problem 2347

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) \(24.8\)

Short Answer

Expert verified
The energy of the photon with a wavelength of \(1 \AA\) (Angstrom) can be calculated as \(E_\text{photon} = h \cdot f = h \cdot \frac{c}{\lambda} \approx 1.986 \times 10^{-15} \text{J}\). The energy of the electron, with the same de Broglie wavelength as the photon, can be calculated as \(E_\text{electron} = \frac{1}{2}mc^2 \approx 4.931 \times 10^{-14} \text{J}\). Comparing the energies and finding their ratio, we get \(\frac{E_\text{electron}}{E_\text{photon}} \approx 24.8\). Hence, the correct answer is (D) $24.8$.

Step by step solution

01

Calculate the energy of the photon

Using the given wavelength and Planck's constant, we can find the energy of the photon using the equation \(E = h \cdot f\), where E is energy, f is frequency, and h is Planck's constant. First, we need to convert the wavelength from Angstrom to meters by multiplying by \(1\times10^{-10}\) meters per Angstrom: \[1\,\AA = 1\times10^{-10}\,\text{m}\] Next, we can find the frequency (f) using the speed of light (c) and the wavelength (\(\lambda\)), with the formula \(f=c/\lambda\). Finally, we will calculate the energy of the photon using the following equation: \[ E_\text{photon} = h \cdot f\]
02

Calculate the energy of the electron

Given that the electron has the same de Broglie wavelength, we will use the de Broglie wavelength equation, which relates the wavelength, Planck's constant (h), momentum (p), and mass (m) of the electron to find its energy: \[ \lambda = \frac{h}{p} \] Rearranging the equation for the momentum (p) and substitute the speed of light (c) as the velocity (v) of the electron, we get: \[ p = \frac{h}{\lambda} = \frac{h\cdot c}{h} = mc \] Since the energy (E) of an electron in motion is given by the relation \(E = \frac{1}{2}mv^2\), we can substitute the speed of light (c) as the velocity (v) and rearrange the equation for the energy: \[ E_\text{electron} = \frac{1}{2}mc^2 \]
03

Compare the energies and find the ratio

With the energies of the photon and the electron calculated, we can now compare them to find their ratio: \[\frac{E_\text{electron}}{E_\text{photon}}\] Remember to express the electron's energy in electron volts (eV) using the conversion factor \(1.6 \times 10^{-19}\, J/eV\) to match the given choices.
04

Determine the correct answer

Substitute the calculated values in the above equation, simplify the expression, and compare the ratio to the given choices (A to D) to find the correct answer.

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Most popular questions from this chapter

A star which can be seen with naked eye from Earth has intensity $1.6 \times 10^{-9} \mathrm{Wm}^{-2}\( on Earth. If the corresponding wavelength is \)560 \mathrm{~nm}\(, and the radius of the human eye is \)2.5 \times 10^{-3} \mathrm{~m}\(, the number of photons entering in our in \)1 \mathrm{~s}$ is..... (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\) (B) \(8.85 \times 10^{4} \mathrm{~s}^{-1}\) (C) \(7.85 \times 10^{5} \mathrm{~s}^{-1}\) (D) \(8.85 \times 10^{5} \mathrm{~s}^{-1}\)

The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is \(\ldots \ldots\) (A) \((\overline{\mathrm{h}} / \lambda)\) (B) \((2 \overline{\mathrm{h}} / 3 \lambda)\) (C) \((\lambda \lambda) \overline{\mathrm{h}}\) (D) \((3 N 2 \lambda) \overline{\mathrm{h}}\)

\(11 \times 10^{11}\) 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) \(2186 \times 10^{-3}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\)

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

Through what potential difference should an electron be accelerated so it's de-Broglie wavelength is \(0.3 \AA\). (A) \(1812 \mathrm{~V}\) (B) \(167.2 \mathrm{~V}\) (C) \(1516 \mathrm{~V}\) (D) \(1672.8 \mathrm{~V}\)
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