Check that the ratio ke2 /Gme mp is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

Physics
Check that the ratio ke2 /Gme mp is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

ScanSolve AI Answer

Step 1: Check that the ratio ke2Gmemp\frac{ke^2}{Gm_e m_p} is dimensionless.

First, determine the dimensions of each constant and quantity: • Dimensions of Coulomb's constant kk: From Coulomb's Law, F=kq1q2r2F = k \frac{q_1 q_2}{r^2}, so k=Fr2q1q2k = \frac{F r^2}{q_1 q_2}. [k]=[MLT2][L2][AT][AT]=[ML3T2][A2T2]=[ML3T4A2][k] = \frac{[MLT^{-2}][L^2]}{[AT][AT]} = \frac{[ML^3 T^{-2}]}{[A^2 T^2]} = [ML^3 T^{-4} A^{-2}] • Dimensions of elementary charge ee: The dimension of charge is [AT][AT]. [e]=[AT][e] = [AT] So, the dimension of e2e^2 is: [e2]=[A2T2][e^2] = [A^2 T^2] • Dimensions of gravitational constant GG: From Newton's Law of Gravitation, F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}, so G=Fr2m1m2G = \frac{F r^2}{m_1 m_2}. [G]=[MLT2][L2][M][M]=[ML3T2][M2]=[M1L3T2][G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = \frac{[ML^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}] • Dimensions of electron mass mem_e and proton mass mpm_p: The dimension of mass is [M][M]. [me]=[M][m_e] = [M] [mp]=[M][m_p] = [M] So, the dimension of mempm_e m_p is: [memp]=[M2][m_e m_p] = [M^2]

Now, substitute these dimensions into the ratio ke2Gmemp\frac{ke^2}{Gm_e m_p}: [ke2Gmemp]=[k][e2][G][memp]\left[\frac{ke^2}{Gm_e m_p}\right] = \frac{[k][e^2]}{[G][m_e m_p]} =[ML3T4A2][A2T2][M1L3T2][M2]= \frac{[ML^3 T^{-4} A^{-2}][A^2 T^2]}{[M^{-1} L^3 T^{-2}][M^2]} Combine the terms in the numerator: [ML3T4A2][A2T2]=[M1L3T4+2A2+2]=[ML3T2A0]=[ML3T2][ML^3 T^{-4} A^{-2}][A^2 T^2] = [M^1 L^3 T^{-4+2} A^{-2+2}] = [ML^3 T^{-2} A^0] = [ML^3 T^{-2}] Combine the terms in the denominator: [M1L3T2][M2]=[M1+2L3T2]=[M1L3T2][M^{-1} L^3 T^{-2}][M^2] = [M^{-1+2} L^3 T^{-2}] = [M^1 L^3 T^{-2}] Substitute these back into the ratio: [ke2Gmemp]=[ML3T2][M1L3T2]\left[\frac{ke^2}{Gm_e m_p}\right] = \frac{[ML^3 T^{-2}]}{[M^1 L^3 T^{-2}]} Simplify the exponents: =[M11L33T2(2)]=[M0L0T0]= [M^{1-1} L^{3-3} T^{-2-(-2)}] = [M^0 L^0 T^0] Since all exponents are zero, the ratio is dimensionless.

Step 2: Determine the value of this ratio. Use the following physical constants: • Coulomb's constant k=8.98755×109Nm2/C2k = 8.98755 \times 10^9 \, N m^2/C^2 • Elementary charge e=1.602176634×1019Ce = 1.602176634 \times 10^{-19} \, \text{C} • Gravitational constant G=6.67430×1011Nm2/kg2G = 6.67430 \times 10^{-11} \, N m^2/kg^2 • Electron mass me=9.1093837015×1031kgm_e = 9.1093837015 \times 10^{-31} \, \text{kg} • Proton mass mp=1.67262192369×1027kgm_p = 1.67262192369 \times 10^{-27} \, \text{kg}

Calculate the numerator ke2ke^2: ke2=(8.98755×109Nm2/C2)×(1.602176634×1019C)2ke^2 = (8.98755 \times 10^9 \, N m^2/C^2) \times (1.602176634 \times 10^{-19} \, C)^2 ke2=(8.98755×109)×(2.5669699×1038)Nm2ke^2 = (8.98755 \times 10^9) \times (2.5669699 \times 10^{-38}) \, N m^2 ke2=2.307077×1028Nm2ke^2 = 2.307077 \times 10^{-28} \, N m^2

Calculate the denominator GmempGm_e m_p:

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