Certainly! Let's determine the de Broglie wavelength of an electron given its kinetic energy of 60 cV (centivolts).
### Step-by-Step Solution:
1. Convert the kinetic energy from cV to Joules:
Since 1 centivolt (cV) is equivalent to [tex]\(1.6 \times 10^{-19}\)[/tex] Joules (as 1 electronvolt (eV) is [tex]\(1.6 \times 10^{-19}\)[/tex] Joules and 1 cV is 0.01 eV), we can convert the kinetic energy:
[tex]\[
\text{Kinetic energy in Joules} = 60 \, \text{cV} \times 1.6 \times 10^{-19} \, \text{J/cV}
\][/tex]
[tex]\[
\text{Kinetic energy in Joules} = 9.6 \times 10^{-18} \, \text{J}
\][/tex]
2. Determine the velocity of the electron:
Using the kinetic energy formula for non-relativistic speeds:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
where [tex]\(m\)[/tex] is the mass of the electron ([tex]\(9.10938356 \times 10^{-31}\)[/tex] kg). Solving for [tex]\(v\)[/tex]:
[tex]\[
v = \sqrt{\frac{2 \times KE}{m}}
\][/tex]
Substituting the known values:
[tex]\[
v = \sqrt{\frac{2 \times 9.6 \times 10^{-18}}{9.10938356 \times 10^{-31}}}
\][/tex]
[tex]\[
v \approx 4590987.601589642 \, \text{m/s}
\][/tex]
3. Calculate the de Broglie wavelength:
The de Broglie wavelength ([tex]\(\lambda\)[/tex]) is given by:
[tex]\[
\lambda = \frac{h}{p}
\][/tex]
where [tex]\(h\)[/tex] is Planck's constant ([tex]\(6.62607015 \times 10^{-34}\)[/tex] Js) and [tex]\(p\)[/tex] is the momentum of the electron ([tex]\(p = m \cdot v\)[/tex]). First, calculate the momentum:
[tex]\[
p = m \times v
\][/tex]
[tex]\[
p = 9.10938356 \times 10^{-31} \times 4590987.601589642
\][/tex]
[tex]\[
p \approx 4.181801772 \times 10^{-24} \, \text{kg m/s}
\][/tex]
Now, calculate the wavelength:
[tex]\[
\lambda = \frac{6.62607015 \times 10^{-34}}{4.181801772 \times 10^{-24}}
\][/tex]
[tex]\[
\lambda \approx 1.5843857242663138 \times 10^{-10} \, \text{m}
\][/tex]
### Conclusion:
The de Broglie wavelength of an electron with a kinetic energy of 60 cV is approximately [tex]\(1.584 \times 10^{-10}\)[/tex] meters.
