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Answer :
To solve the problem, we need to determine the frequency of a helium ion that is accelerated by a voltage of 3700 V. The ion has a charge [tex]\( Q = +2e \)[/tex] and a mass of [tex]\( 6.6 \times 10^{-27} \)[/tex] kg.
Here are the steps to find the frequency:
1. Determine the charge (Q) in coulombs (C):
The charge of the helium ion is twice the elementary charge [tex]\( e \)[/tex]:
[tex]\[
Q = 2 \times 1.60219 \times 10^{-19} \, \text{C}
\][/tex]
[tex]\[
Q = 3.20438 \times 10^{-19} \, \text{C}
\][/tex]
2. Calculate the kinetic energy (K.E.) gained by the ion:
When the ion is accelerated by a voltage [tex]\( V \)[/tex], the kinetic energy is given by:
[tex]\[
\text{K.E.} = Q \times V
\][/tex]
Substituting the given voltage:
[tex]\[
\text{K.E.} = 3.20438 \times 10^{-19} \, \text{C} \times 3700 \, \text{V}
\][/tex]
[tex]\[
\text{K.E.} = 1.1856206 \times 10^{-15} \, \text{J}
\][/tex]
3. Find the velocity (v) of the ion:
The kinetic energy is also expressed as:
[tex]\[
\text{K.E.} = \frac{1}{2} \times \text{mass} \times \text{velocity}^2
\][/tex]
Solving for velocity:
[tex]\[
\text{velocity} = \sqrt{\frac{2 \times \text{K.E.}}{\text{mass}}}
\][/tex]
Plugging in the values:
[tex]\[
\text{velocity} = \sqrt{\frac{2 \times 1.1856206 \times 10^{-15}}{6.6 \times 10^{-27}}}
\][/tex]
[tex]\[
\text{velocity} \approx 599398.8402532738 \, \text{m/s}
\][/tex]
4. Calculate the De Broglie wavelength (λ):
The De Broglie wavelength is given by:
[tex]\[
\lambda = \frac{h}{\text{mass} \times \text{velocity}}
\][/tex]
where [tex]\( h \)[/tex] is Planck's constant [tex]\( 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]:
[tex]\[
\lambda = \frac{6.62607015 \times 10^{-34}}{6.6 \times 10^{-27} \times 599398.8402532738}
\][/tex]
[tex]\[
\lambda \approx 1.6749282035698589 \times 10^{-13} \, \text{m}
\][/tex]
5. Determine the radius of the circular path:
For simplicity, assume the ion travels in a circular path with a radius approximated by:
[tex]\[
\text{radius} = \frac{\lambda}{2\pi}
\][/tex]
[tex]\[
\text{radius} = \frac{1.6749282035698589 \times 10^{-13}}{2 \times 3.141592653589793}
\][/tex]
[tex]\[
\text{radius} \approx 2.6657310292217138 \times 10^{-14} \, \text{m}
\][/tex]
6. Calculate the frequency (f):
The frequency of motion in a circular path is given by:
[tex]\[
f = \frac{\text{velocity}}{2\pi \times \text{radius}}
\][/tex]
[tex]\[
f = \frac{599398.8402532738}{2 \times 3.141592653589793 \times 2.6657310292217138 \times 10^{-14}}
\][/tex]
[tex]\[
f \approx 3.5786539326028713 \times 10^{18} \, \text{Hz}
\][/tex]
Therefore, the frequency of the helium ion is approximately [tex]\( 3.5786539326028713 \times 10^{18} \)[/tex] Hz.
Here are the steps to find the frequency:
1. Determine the charge (Q) in coulombs (C):
The charge of the helium ion is twice the elementary charge [tex]\( e \)[/tex]:
[tex]\[
Q = 2 \times 1.60219 \times 10^{-19} \, \text{C}
\][/tex]
[tex]\[
Q = 3.20438 \times 10^{-19} \, \text{C}
\][/tex]
2. Calculate the kinetic energy (K.E.) gained by the ion:
When the ion is accelerated by a voltage [tex]\( V \)[/tex], the kinetic energy is given by:
[tex]\[
\text{K.E.} = Q \times V
\][/tex]
Substituting the given voltage:
[tex]\[
\text{K.E.} = 3.20438 \times 10^{-19} \, \text{C} \times 3700 \, \text{V}
\][/tex]
[tex]\[
\text{K.E.} = 1.1856206 \times 10^{-15} \, \text{J}
\][/tex]
3. Find the velocity (v) of the ion:
The kinetic energy is also expressed as:
[tex]\[
\text{K.E.} = \frac{1}{2} \times \text{mass} \times \text{velocity}^2
\][/tex]
Solving for velocity:
[tex]\[
\text{velocity} = \sqrt{\frac{2 \times \text{K.E.}}{\text{mass}}}
\][/tex]
Plugging in the values:
[tex]\[
\text{velocity} = \sqrt{\frac{2 \times 1.1856206 \times 10^{-15}}{6.6 \times 10^{-27}}}
\][/tex]
[tex]\[
\text{velocity} \approx 599398.8402532738 \, \text{m/s}
\][/tex]
4. Calculate the De Broglie wavelength (λ):
The De Broglie wavelength is given by:
[tex]\[
\lambda = \frac{h}{\text{mass} \times \text{velocity}}
\][/tex]
where [tex]\( h \)[/tex] is Planck's constant [tex]\( 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]:
[tex]\[
\lambda = \frac{6.62607015 \times 10^{-34}}{6.6 \times 10^{-27} \times 599398.8402532738}
\][/tex]
[tex]\[
\lambda \approx 1.6749282035698589 \times 10^{-13} \, \text{m}
\][/tex]
5. Determine the radius of the circular path:
For simplicity, assume the ion travels in a circular path with a radius approximated by:
[tex]\[
\text{radius} = \frac{\lambda}{2\pi}
\][/tex]
[tex]\[
\text{radius} = \frac{1.6749282035698589 \times 10^{-13}}{2 \times 3.141592653589793}
\][/tex]
[tex]\[
\text{radius} \approx 2.6657310292217138 \times 10^{-14} \, \text{m}
\][/tex]
6. Calculate the frequency (f):
The frequency of motion in a circular path is given by:
[tex]\[
f = \frac{\text{velocity}}{2\pi \times \text{radius}}
\][/tex]
[tex]\[
f = \frac{599398.8402532738}{2 \times 3.141592653589793 \times 2.6657310292217138 \times 10^{-14}}
\][/tex]
[tex]\[
f \approx 3.5786539326028713 \times 10^{18} \, \text{Hz}
\][/tex]
Therefore, the frequency of the helium ion is approximately [tex]\( 3.5786539326028713 \times 10^{18} \)[/tex] Hz.
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