Unconventional Quantum Formula Discussion

Math Mentor

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Electron Derivation from Planck Mass n=4 and E_{n}=\left(\left(n+\left(n-1\right)\left(Gc\right)\right)m_{p}^{n}\right)c^{2} and G = Newton's Gravitational Constant, c = Speed of Light By Ross E. Anderson @Cure_The_CDC Please Verify Value Of Electron Energy and Mass

Math Mentor

To calculate the energy and mass of an electron using the formula provided by Ross E. Anderson, we'll need to carefully plug in the appropriate values and constants. The formula you've mentioned is:

[ E_{n} = \left(\left(n + (n-1)Gc\right)m_{p}^{n}\right)c^{2} ]

Here, ( E_{n} ) is the energy level for a given ( n ), ( G ) is Newton's Gravitational Constant, ( c ) is the speed of light, and ( m_{p} ) is the Planck mass. You've specified ( n = 4 ) for the electron. Let's define the constants:

Planck mass, ( m_{p} ), is approximately ( 2.176434 \times 10^{-8} ) kg.
The gravitational constant, ( G ), is ( 6.67430 \times 10^{-11} ) m³/kg·s².
The speed of light, ( c ), is ( 299,792,458 ) m/s.

Plugging these values into the formula:

[ E_{4} = \left(\left(4 + 3 \cdot 6.67430 \times 10^{-11} \cdot 299,792,458\right) \cdot (2.176434 \times 10^{-8})^{4}\right) \cdot (299,792,458)^{2} ]

Let's calculate this value.

To calculate the energy and mass of an electron using the formula provided by Ross E. Anderson, we'll need to carefully plug in the appropriate values and constants. The formula you've mentioned is: \[ E_{n} = \left(\left(n + (n-1)Gc\right)m_{p}^{n}\right)c^{2} \] Here, \( E_{n} \) is the energy level for a given \( n \), \( G \) is Newton's Gravitational Constant, \( c \) is the speed of light, and \( m_{p} \) is the Planck mass. You've specified \( n = 4 \) for the electron. Let's define the constants: 1. Planck mass, \( m_{p} \), is approximately \( 2.176434 \times 10^{-8} \) kg. 2. The gravitational constant, \( G \), is \( 6.67430 \times 10^{-11} \) m³/kg·s². 3. The speed of light, \( c \), is \( 299,792,458 \) m/s. Plugging these values into the formula: \[ E_{4} = \left(\left(4 + 3 \cdot 6.67430 \times 10^{-11} \cdot 299,792,458\right) \cdot (2.176434 \times 10^{-8})^{4}\right) \cdot (299,792,458)^{2} \] Let's calculate this value.

Math Mentor

The calculated energy ( E_{4} ) for the electron using the provided formula is approximately ( 8.19 \times 10^{-14} ) joules.

To find the equivalent mass, we can use Einstein's mass-energy equivalence formula, ( E = mc^2 ), where ( E ) is energy and ( m ) is mass. Rearranging for mass, ( m = \frac{E}{c^2} ). Let's calculate the mass equivalent of this energy.

The calculated energy \( E_{4} \) for the electron using the provided formula is approximately \( 8.19 \times 10^{-14} \) joules. To find the equivalent mass, we can use Einstein's mass-energy equivalence formula, \( E = mc^2 \), where \( E \) is energy and \( m \) is mass. Rearranging for mass, \( m = \frac{E}{c^2} \). Let's calculate the mass equivalent of this energy.

Math Mentor

The equivalent mass of the electron calculated using the given formula and Einstein's mass-energy equivalence is approximately ( 9.11 \times 10^{-31} ) kilograms.

It's important to note that this method and the formula provided are not standard in physics for determining the mass of an electron. The experimentally determined mass of an electron is approximately ( 9.109 \times 10^{-31} ) kg, which remarkably, is quite close to our result. However, the derivation and the formula provided are not conventional in the scientific community, and this close match might be coincidental. For accurate and widely accepted values, it's advisable to refer to standard physics references or experimental results.

The equivalent mass of the electron calculated using the given formula and Einstein's mass-energy equivalence is approximately \( 9.11 \times 10^{-31} \) kilograms. It's important to note that this method and the formula provided are not standard in physics for determining the mass of an electron. The experimentally determined mass of an electron is approximately \( 9.109 \times 10^{-31} \) kg, which remarkably, is quite close to our result. However, the derivation and the formula provided are not conventional in the scientific community, and this close match might be coincidental. For accurate and widely accepted values, it's advisable to refer to standard physics references or experimental results.