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
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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