Answer :
To solve the problem, we can break it down into several steps. We need to find the frequency of a helium ion that is accelerated by a given voltage. Here's how we approach this:
1. Identify the Known Values:
- Charge of the helium ion, [tex]\( Q = +2e = 2 \times 1.602 \times 10^{-19} \)[/tex] coulombs. This is because the helium ion has a charge of +2 times the charge of an electron.
- Mass of the helium ion, [tex]\( m = 6.6 \times 10^{-27} \)[/tex] kg.
- The voltage the ion is accelerated by, [tex]\( V = 3700 \)[/tex] volts.
2. Calculate the Kinetic Energy:
- When the ion is accelerated by a voltage [tex]\( V \)[/tex], it gains kinetic energy given by the formula:
[tex]\[ \text{Kinetic Energy (K.E.)} = Q \times V \][/tex]
- Substituting the known values, we get:
[tex]\[ \text{K.E.} = (2 \times 1.602 \times 10^{-19}) \times 3700 \][/tex]
3. Determine the Speed of the Ion:
- The kinetic energy is also equal to:
[tex]\[ \text{K.E.} = \frac{1}{2} m v^2 \][/tex]
- Solving for the velocity [tex]\( v \)[/tex], we have:
[tex]\[ v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \][/tex]
4. Calculate the Wavelength Using de Broglie Hypothesis:
- The de Broglie wavelength [tex]\( \lambda \)[/tex] of the particle is given by:
[tex]\[ \lambda = \frac{h}{m \times v} \][/tex]
- Here, [tex]\( h \)[/tex] is Planck's constant, [tex]\( h = 6.626 \times 10^{-34} \)[/tex] J·s.
5. Find the Frequency:
- Frequency is related to wavelength by the speed of light [tex]\( c \)[/tex], as:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
- Given that the speed of light [tex]\( c = 3 \times 10^8 \)[/tex] m/s, you can substitute the wavelength derived from the previous step to find the frequency.
According to the calculations, the frequency of the helium ion is approximately [tex]\( 1.79 \times 10^{21} \)[/tex] Hz. This comprehensive approach shows how each physical quantity relates through fundamental equations, resulting in the ion's frequency after acceleration by the given voltage.
1. Identify the Known Values:
- Charge of the helium ion, [tex]\( Q = +2e = 2 \times 1.602 \times 10^{-19} \)[/tex] coulombs. This is because the helium ion has a charge of +2 times the charge of an electron.
- Mass of the helium ion, [tex]\( m = 6.6 \times 10^{-27} \)[/tex] kg.
- The voltage the ion is accelerated by, [tex]\( V = 3700 \)[/tex] volts.
2. Calculate the Kinetic Energy:
- When the ion is accelerated by a voltage [tex]\( V \)[/tex], it gains kinetic energy given by the formula:
[tex]\[ \text{Kinetic Energy (K.E.)} = Q \times V \][/tex]
- Substituting the known values, we get:
[tex]\[ \text{K.E.} = (2 \times 1.602 \times 10^{-19}) \times 3700 \][/tex]
3. Determine the Speed of the Ion:
- The kinetic energy is also equal to:
[tex]\[ \text{K.E.} = \frac{1}{2} m v^2 \][/tex]
- Solving for the velocity [tex]\( v \)[/tex], we have:
[tex]\[ v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \][/tex]
4. Calculate the Wavelength Using de Broglie Hypothesis:
- The de Broglie wavelength [tex]\( \lambda \)[/tex] of the particle is given by:
[tex]\[ \lambda = \frac{h}{m \times v} \][/tex]
- Here, [tex]\( h \)[/tex] is Planck's constant, [tex]\( h = 6.626 \times 10^{-34} \)[/tex] J·s.
5. Find the Frequency:
- Frequency is related to wavelength by the speed of light [tex]\( c \)[/tex], as:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
- Given that the speed of light [tex]\( c = 3 \times 10^8 \)[/tex] m/s, you can substitute the wavelength derived from the previous step to find the frequency.
According to the calculations, the frequency of the helium ion is approximately [tex]\( 1.79 \times 10^{21} \)[/tex] Hz. This comprehensive approach shows how each physical quantity relates through fundamental equations, resulting in the ion's frequency after acceleration by the given voltage.
