An oxygen sensor is being tested with a digital multimeter using the MIN/MAX function. The readings are: minimum = 78 mV; maximu
1 answer:
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Answer:
de Broglie wavelength of an electron with speed 0.78 c taking relativistic effects into account is given as:
λ = 1.943 * 10^(-12) m
Explanation:
Given:
v = 0.78 c
we know:
c = speed of light = 3 * 10^8 m/s
mass of electron = m = 9.1 × 10-31 kg
de Broglie wavelength:
In 1924 a French physicist Louis de Broglie assumed that for particles the same relations are valid as for the photon:
(Dual-nature of a particle)
Let the wavelength be = λ
According to de Broglie:
λ = h/p = h/mv
where h is planck's constant = 6.626176 x 10^-34 Js
and p is momentum.
Taking relativistic effects into account, we know that the momentum of the particle changes by a factor 'γ'.
At low speed, γ is almost 1. However, at very high velocity (comparable to light), it has a great effect on momentum.
γ =
γ = 1.6
Now at 0.78 c, considering relativistic effects, we know:
λ = h/γp = h/γ*mv
= (6.62 x 10^(-34))/(1.6*0.78*3*10^(8)*9.1 × 10-31
λ = 1.943 * 10^(-12) m
Answer:
Electric potential energy at the negative terminal:
Explanation:
When a particle with charge travels across a potential difference
, then its change in electric potential energy is
In this problem, we know that:
The particle is an electron, so its charge is
We also know that the positive terminal is at potential
While the negative terminal is at potential
Therefore, the potential difference (final minus initial) is
So, the change in potential energy of the electron is
This means that the electron when it is at the negative terminal has of energy more than when it is at the positive terminal.
Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is