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What is the De Broglie wavelength associated with an electron moving under a potential difference of ${\text{1}}{{\text{0}}^4}$ V(A) 12.27nm(B) 1nm(C) 0.01277 nm(D) 0.1227 nm

Last updated date: 19th Sep 2024
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HintDe Broglie wavelength of a particle is obtained by finding the ratio of Planck's constant wrt the momentum. Momentum is obtained by $\sqrt {2 \times m \times KE}$. And this momentum is used to obtain the De Broglie wavelength

Complete step-by-step solution
De Broglie stated that every moving particle will have a wavelength, which could even be so less that they are not felt or are treated negligible. These matter waves are given by the equation,
$\lambda = \dfrac{h}{{mv}}$
When an electron is accelerated by a potential of V, it gains some velocity or momentum. This momentum is given by
KE = energy gained through potential difference
As we know that KE can be expressed in terms of momentum as,
$\dfrac{{{p^2}}}{{2m}}{\text{ }} = {\text{ }}eV \\ p = \sqrt {2meV} \\ p = \sqrt {2x9.1x{{10}^{ - 31}}x1.6x{{10}^{ - 19}}x10000} \\ p = 5.396x{10^{ - 23}} \\$
Substituting the value of p in the de Broglie equation,
$\lambda = \dfrac{{6.626x{{10}^{ - 34}}}}{{5.396x{{10}^{ - 23}}}} = 1.22x{10^{ - 11}} \\ \lambda = 0.0122x{10^{ - 9}} = 0.0122nm \\$

Therefore the correct answer is option C

Note Matter waves are formed whenever an object moves, but it is difficult to observe them most of the time. Let’s say that we have a body $m = 1$ kg and it is moving with velocity $= 10m/s$. The obtained De Broglie wavelength = $\lambda = .6 \ *{10^{ - 33}}m$. This length is not just invisible to eyes but also invisible to all the devices in the world right now in 2020