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  • What is the de Broglie wavelength of an electron that is accelerated through a potential difference of 100 V? (Take the mass of an electron as <img alt="9.1\times 10^{-31}" src="https://entran

What is the de Broglie wavelength of an electron that is accelerated through a potential difference of 100 V? (Take the mass of an electron as 9.1\times 10^{-31} kg and the value of Planck's constant as 6.626\times 10^{-34} Js.)

Option: 1

1.138\times 10^{-10}m


Option: 2

1.095 \times 10^{-7}m.


Option: 3

1.095 \times 10^{-17}m.


Option: 4

1.095 \times 10^{-11}m.


Answers (1)

best_answer

To calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V, we need to use the following formula:
\lambda= \frac{h}{p}
where \lambda is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the electron. We can find the momentum of the electron by using the following equation:
p= \sqrt{2mK}
where m is the mass of the electron, K is the kinetic energy of the electron. We can find the kinetic energy of the electron using the potential difference and the charge of the electron:
K= eV

where e is the charge of the electron and V is the potential difference.
Putting it all together, we get:
K=e V=\left(1.602 \times 10^{-19} \mathrm{C}\right) \times(100 \mathrm{~V})=1.602 \times 10^{-17} \mathrm{~J}
p=\sqrt{2 m K}=\sqrt{2 \times 9.109 \times 10^{-31} \mathrm{~kg} \times 1.602 \times 10^{-17} \mathrm{~J}}=5.821 \times 10^{-24} \mathrm{kgm} / \mathrm{s}
\lambda=\frac{h}{p}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{5.821 \times 10^{-24} \mathrm{kgm} / \mathrm{s}}=1.138 \times 10^{-10} \mathrm{~m}

Therefore, the de Broglie wavelength of an electron accelerated through a potential difference of 100 V is approximately 1.138\times 10^{-10}m.

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jitender.kumar

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