Question

An evacuated tube uses an accelerating voltage of 40 kv to accelerate electrons to hit a copper plate and produce x rays. non-relativistically, what would be the maximum speed of these electrons?

Answer 1

The energy acquired by the electrons due to the accelerating voltage is given by:
$\Delta U=e \Delta V$
where
$e=1.6 \cdot 10^{-19} C$ is the electron charge
$\Delta V=40 kV=4 \cdot 10^4 V$ is the accelerating voltage
Substituting the numbers into the formula, we find
$\Delta U=(1.6 \cdot 10^{-19}C)(4 \cdot 10^4 V)=6.4 \cdot 10^{-15} J$

If the electrons start from rest, this value corresponds to the final kinetic energy of the electrons:
$K_f = \Delta U$
Keeping in mind the equation for the kinetic energy:
$K_f = \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of one electron, we can rearrange the formula to calculate the final speed of the electrons:
$v= \sqrt{ \frac{2 K_f }{m} } = \sqrt{ \frac{2 \cdot 6.4 \cdot 10^{-15}J}{9.1 \cdot 10^{-31} kg} } =1.19 \cdot 10^8 m/s$

Answer 2

The speed at which the electrons hit the copper plate due to the accelerating voltage is $\boxed{1.19 \times {{10}^8}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Further Explanation:

The potential difference, through which the electron, provides the potential energy and this potential energy provided by the potential difference to the electron leads to the increase in the kinetic energy of the electron.

The charge on the electron is $1.6 \times {10^{ - 19}}\,{\text{C}}$ and the mass of the electron is $9.1 \times {10^{ - 31}}\,{\text{kg}}$.

The expression for the conservation of the energy of the electron in the potential difference is:

$\Delta PE = \Delta KE$  

Substitute $q\Delta V$ for $\Delta PE$ and $\dfrac{1}{2}m{v^2}$ for $\Delta KE$ in above expression.

$\begin{aligned}q\Delta V &= \frac{1}{2}m{v^2} \hfill \\v&= \sqrt {\frac{{2q\Delta V}}{m}} \hfill \\\end{aligned}$  

Here, $\Delta V$ is the potential difference, $m$ is the mass of the ion, $q$ is the charge over the electron.

Substitute $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $q$, $9.1 \times {10^{ - 31}}\,{\text{kg}}$ for $m$ and $4 \times {10^4}\,{\text{V}}$ for $V$ in above expression.

$\begin{aligned}v&= \sqrt {\frac{{2 \times \left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)\left( {4 \times {{10}^4}\,{\text{V}}} \right)}}{{9.1 \times {{10}^{ - 31}}\,{\text{kg}}}}}\\&= 1.19 \times {10^8}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$  

Thus, the speed at which the electrons hit the copper plate due to the accelerating voltage is $\boxed{1.19 \times {{10}^8}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords: Potential energy, potential difference, accelerate, kinetic energy, speed, electron, charge, significant figures, hits, copper plate, non-relativistically.