An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this... An electron is accelerated through a potential difference of 64 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? What is the wavelength of an electron with kinetic energy 49.8 eV?
Understand the Problem
The question is asking for the de-Broglie wavelength associated with an electron accelerated through a potential difference of 64 volts and to identify the part of the electromagnetic spectrum. It also asks for the wavelength of an electron with a kinetic energy of 49.8 electron volts.
Answer
The de-Broglie wavelength for 64 V is approximately $1.94 \times 10^{-12} \text{ m}$ (gamma rays); for 49.8 eV, it is approximately $1.65 \times 10^{-10} \text{ m}$.
Answer for screen readers
The de-Broglie wavelength associated with an electron accelerated through a potential difference of 64 volts is approximately $1.94 \times 10^{-12} \text{ m}$, which corresponds to the gamma ray part of the electromagnetic spectrum. The wavelength for an electron with kinetic energy of 49.8 eV is approximately $1.65 \times 10^{-10} \text{ m}$.
Steps to Solve
- Calculate the kinetic energy of the electron The kinetic energy (KE) of an electron accelerated through a potential difference (V) can be found using the equation:
$$ KE = eV $$
where $e$ is the charge of the electron ($1.6 \times 10^{-19} \text{ C}$) and $V = 64 \text{ V}$.
Substituting in the values gives:
$$ KE = (1.6 \times 10^{-19} , \text{C})(64 , \text{V}) = 1.024 \times 10^{-17} , \text{J} $$
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Convert kinetic energy to eV To express this energy in electron volts (already done), we note $64 \text{ V} = 64 \text{ eV}$.
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Calculate the de-Broglie wavelength The de-Broglie wavelength ($\lambda$) is related to momentum ($p$) with the formula:
$$ \lambda = \frac{h}{p} $$
For an electron, momentum can be calculated from kinetic energy:
$$ p = \sqrt{2m_e KE} $$
where $m_e$ is the mass of the electron ($9.11 \times 10^{-31} , \text{kg}$).
So:
- Calculate momentum:
$$ p = \sqrt{2(9.11 \times 10^{-31} , \text{kg})(1.024 \times 10^{-17} , \text{J})} $$
- Insert $p$ into the de-Broglie wavelength equation:
Using Planck's constant $h = 6.63 \times 10^{-34} , \text{Js}$:
$$ \lambda = \frac{h}{p} $$
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Calculate for 49.8 eV Follow the same process to calculate the de-Broglie wavelength for an electron with a kinetic energy of 49.8 eV:
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Convert $49.8 , \text{eV}$ to joules:
$$ KE = 49.8 , \text{eV} \times 1.6 \times 10^{-19} , \text{J/eV} = 7.968 \times 10^{-18} , \text{J} $$
- Calculate momentum again as in the previous formula and then:
$$ p = \sqrt{2m_e KE} $$
- Substitute momentum back into the de-Broglie wavelength formula.
The de-Broglie wavelength associated with an electron accelerated through a potential difference of 64 volts is approximately $1.94 \times 10^{-12} \text{ m}$, which corresponds to the gamma ray part of the electromagnetic spectrum. The wavelength for an electron with kinetic energy of 49.8 eV is approximately $1.65 \times 10^{-10} \text{ m}$.
More Information
The de-Broglie wavelength represents the wave-like nature of particles. The calculated wavelengths place electrons in the range of gamma rays and X-rays, illustrating their significance in quantum mechanics and particle physics.
Tips
- Forgetting to convert electron volts to joules when calculating kinetic energy.
- Misapplying the de-Broglie wavelength formula by using incorrect values for mass or momentum.
- Not understanding the significance of wavelength in the electromagnetic spectrum, leading to misclassification.
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