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Chapter 17: Problem 2333

If we take accelerating voltage \(\mathrm{V}=50 \mathrm{~V}\), electric charge of electron \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) and mass of electron \(\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}\) find the wavelength of concerned election. (A) \(0.1735 \mathrm{~A}\) (B) \(1.735 \AA\) (C) \(17.35 \AA\) (D) \(1735 \AA\)

Short Answer

Expert verified
The short answer is (B) \(1.735 \AA\).

Step by step solution

01

Write down the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the electron.
02

Write down the energy relation for an electron in an accelerating voltage

The energy relation for an electron in an accelerating voltage is given by: \[ E = eV \] where \(E\) is the energy of the electron, \(e\) is the electric charge of the electron, and \(V\) is the accelerating voltage.
03

Relate the energy and the momentum of the electron

The energy and the momentum of the electron are related by the following formula: \[ E = \frac{p^2}{2m} \] where \(E\) is the energy, \(p\) is the momentum, and \(m\) is the mass of the electron.
04

Express the momentum in terms of energy, electric charge, and accelerating voltage

Using the energy relation and the momentum-energy relation, we can express the momentum of the electron as follows: \[ p = \sqrt{2m eV} \]
05

Calculate the de Broglie wavelength using the given values

Now, we can substitute the given values and the derived momentum expression into the de Broglie wavelength formula to calculate the wavelength: \[ \lambda = \frac{h}{\sqrt{2m eV}} \] Plugging in the values, \(h = 6.626 \times 10^{-34} \mathrm{Js}\), \(V = 50 \mathrm{~V}\), \(e = 1.6 \times 10^{-19} \mathrm{C}\), and \(m = 9.1 \times 10^{-31} \mathrm{~kg}\): \[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2(9.1 \times 10^{-31})(1.6 \times 10^{-19})(50)}} \]
06

Solve for the wavelength and convert it to Angstroms

Now, solving the above expression, we get the following value for the wavelength in meters: \[ \lambda = 1.735 \times 10^{-10} \mathrm{m} \] To convert it to Angstroms, we multiply by \(10^{10}\): \[ \lambda = 1.735 \times 10^{10} \times 10^{-10} \mathrm{A} = 1.735\ \mathrm{ A} \] The answer is (B) \(1.735 \AA\).

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Most popular questions from this chapter

An electron is at a distance of \(10 \mathrm{~m}\) form a charge of $10 \mathrm{C}\(. Its total energy is \)15.6 \times 10^{-10} \mathrm{~J}$. Its de Broglie wavelength at this point is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} . \mathrm{K}=9 \times 10^{9} \mathrm{SI}\right)$ (A) \(9.87 \AA\) (B) \(9.87\) Fermi (C) \(8.97 \mathrm{~A}\) (D) \(8.97\) Fermi

The ration of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature \(27^{\circ}\) and \(127^{\circ} \mathrm{C}\) respectively is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) \(\sqrt{(3 / 8)}\) (C) \(\sqrt{(8 / 3)}\) (D) 1

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

An electric bulb of \(100 \mathrm{w}\) converts \(3 \%\) of electrical energy into light energy. If the wavelength of light emitted is \(6625 \AA\), the number of photons emitted is \(1 \mathrm{~s}\) is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(10^{17}\) (B) \(10^{19}\) (C) \(10^{21}\) (D) \(10^{15}\)

The work function of metal is \(5.3 \mathrm{eV}\). What is threshold frequency? (A) \(3.1 \times 10^{15} \mathrm{~Hz}\) (B) \(3.1 \times 10^{45} \mathrm{~Hz}\) (C) \(1.3 \times 10^{15} \mathrm{~Hz}\) (D) \(1.3 \times 10^{45} \mathrm{~Hz}\)
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