
Write the decreasing order of the value of neutron, electron and proton.
Answer
483.3k+ views
Hint: J.J Thomson measures the $\dfrac{e}{m}$ ratio of an electron, it is the ratio of charge to the mass of an electron. Here $m$ is the mass of the particle and $e$ is its energy which is measured in energy.
Complete step by step answer:
In the given question first we have to find the $\dfrac{e}{m}$ value of neutron, electron and proton before writing them in decreasing order. For calculating the value of $\dfrac{e}{m}$ , we required the value of $e$ and $m$ of these particles. Here is the table which has value of $e$ and $m$ :
Now use this value for calculating the $\dfrac{e}{m}$ ratio of Neutron, Electron and proton. Let's find the value one by one.
$\dfrac{e}{m}$ Value of Neutron is $\dfrac{e}{m} = \dfrac{0}{{1.675 \times {{10}^{ - 27}}}} = 0$
or $\dfrac{e}{m}$ value of electron is \[\dfrac{e}{m} = \dfrac{{1.602 \times {{10}^{ - 19}}}}{{9.109 \times {{10}^{ - 31}}}} = 0.1758 \times {10^{12}}\]
and, $\dfrac{e}{m}$ value of proton is $\dfrac{e}{m} = \dfrac{{1.602 \times {{10}^{ - 19}}}}{{1.672 \times {{10}^{ - 27}}}} = 0.9581 \times {10^8}$
Thus, we get the following value of $\dfrac{e}{m}$ the value of electron is the highest among all, then followed by protons and neutron
Therefore the decreasing order of $\dfrac{e}{m}$ value of neutron, electron and proton is
$Electron > proton > Neutron$
Additional information:
A British physicist J.J Thomson calculated the ratio of electrical energy to the mass of cathode ray particles by using specially designed cathode ray tubes. He related the experiment with cathode rays and different energies and strengths of magnetic or electric fields. He was able to calculate the value of $\dfrac{e}{m}$ .
Note:
The charge to mass ratio $\left( {\dfrac{e}{m}} \right)$ quantity is generally useful only for objects that may be treated as particles. The ratio of electrostatic to gravitational forces between two particles will be proportional to the product of their charge to mass ratios.
Complete step by step answer:
In the given question first we have to find the $\dfrac{e}{m}$ value of neutron, electron and proton before writing them in decreasing order. For calculating the value of $\dfrac{e}{m}$ , we required the value of $e$ and $m$ of these particles. Here is the table which has value of $e$ and $m$ :
Subatomic particles | Mass | Charge |
Neutron | $1.672 \times {10^{ - 27}}$ | 0 |
Electron | $9.109 \times {10^{ - 31}}$ | $1.602 \times {10^{ - 19}}$ |
Proton | $1.675 \times {10^{ - 27}}$ | $1.602 \times {10^{ - 19}}$ |
Now use this value for calculating the $\dfrac{e}{m}$ ratio of Neutron, Electron and proton. Let's find the value one by one.
$\dfrac{e}{m}$ Value of Neutron is $\dfrac{e}{m} = \dfrac{0}{{1.675 \times {{10}^{ - 27}}}} = 0$
or $\dfrac{e}{m}$ value of electron is \[\dfrac{e}{m} = \dfrac{{1.602 \times {{10}^{ - 19}}}}{{9.109 \times {{10}^{ - 31}}}} = 0.1758 \times {10^{12}}\]
and, $\dfrac{e}{m}$ value of proton is $\dfrac{e}{m} = \dfrac{{1.602 \times {{10}^{ - 19}}}}{{1.672 \times {{10}^{ - 27}}}} = 0.9581 \times {10^8}$
Thus, we get the following value of $\dfrac{e}{m}$ the value of electron is the highest among all, then followed by protons and neutron
Therefore the decreasing order of $\dfrac{e}{m}$ value of neutron, electron and proton is
$Electron > proton > Neutron$
Additional information:
A British physicist J.J Thomson calculated the ratio of electrical energy to the mass of cathode ray particles by using specially designed cathode ray tubes. He related the experiment with cathode rays and different energies and strengths of magnetic or electric fields. He was able to calculate the value of $\dfrac{e}{m}$ .
Note:
The charge to mass ratio $\left( {\dfrac{e}{m}} \right)$ quantity is generally useful only for objects that may be treated as particles. The ratio of electrostatic to gravitational forces between two particles will be proportional to the product of their charge to mass ratios.
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