The de Broglie wavelength for an electron that has passed the accelerating potential

| June 17, 2021 | education

The de Broglie wavelength for an electron that has passed the accelerating potential difference from a state of rest is 121 pm. the potential difference is?

Data: λ – de Broglie wavelength (λ = 121 pm = 121 * 10-12 m).

Const: h – Planck’s constant (h = 6.62607 * 10 ^ -34 J * s); qе – electron charge (qе = -1.6 * 10 ^ -19 C); me is the mass of an electron (me = 9.11 * 10 ^ -31 kg).

To find out the potential difference that the presented electron has passed, consider the equality: qе * Δφ = р ^ 2 / 2me = (h / λ ^ 2) / 2me = h ^ 2 / (λ ^ 2 * 2me), from where we express: Δφ = h ^ 2 / (λ ^ 2 * 2me * qе).

Let’s perform the calculation: Δφ = h ^ 2 / (λ ^ 2 * 2me * qе) = (6.62607 * 10-34) ^ 2 / ((121 * 10 ^ -12) ^ 2 * 2 * 9.11 * 10 ^ -31 * 1.6 * 10 ^ -19) ≈ 102.9 V.

Answer: The presented electron, according to the calculation, has passed the potential difference of 102.9 V.



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