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The de Broglie wavelength associated with an electron acceleratedthrough a potential difference of 121 V is

Question

The de Broglie wavelength associated with an electron accelerated through a potential difference of 121 V is

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Solution

The de Broglie wavelength (λ) of a particle can be calculated using the de Broglie equation:

λ = h / p

where h is Planck's constant (6.626 x 10^-34 Js) and p is the momentum of the particle.

The momentum of an electron accelerated through a potential difference can be calculated using the equation:

p = sqrt(2meV)

where me is the mass of the electron (9.11 x 10^-31 kg), and V is the potential difference (121 V in this case).

First, calculate the momentum:

p = sqrt(2 * 9.11 x 10^-31 kg * 121 V) p = 1.644 x 10^-24 kg m/s

Then, substitute p into the de Broglie equation to find the wavelength:

λ = 6.626 x 10^-34 Js / 1.644 x 10^-24 kg m/s λ = 4.03 x 10^-10 m

So, the de Broglie wavelength associated with an electron accelerated through a potential difference of 121 V is approximately 4.03 x 10^-10 m.

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