Question

in hydrogen atom the average distance r between the orbital electron and the nuclear proton is 0.529 Å. calculate (1) the magnitude of the electrostatic force (2) the ratio of the electrostatic to the gravitational forces between the two particles
[electronic charge e = 1.6 × 10to the power -19 C, electron mass m e = 9.11×10to the power -31kg proton mass MP = 1.67×10 to the power -27 kg G = 6.67× 10 to the power -11 N.m²/kg²]​

Answer 1

Answer:

in hydrogen atom the average distance r between the orbital electron and the nuclear proton ... - did not match any documents.

Suggestions:

Make sure that all words are spelled correctly.

Try different keywords.

Try more general keywords.

Try fewer keywords.

Answer 2

Answer:

${ \red{ \huge{ \mathfrak{given}}}}$

r = 0.529 Å = 5.29 × 10-¹¹ m,

e = 1.6 × 10-¹⁹ C

me = 9.11 × 10-³¹ kg

mp = 1.67 × 10-²⁷ kg 1/4πe0 = 9 × 10⁹ N•m²/C²

G = 6.67 × 10-¹¹ N•m²/kg²

${ \red{ \huge{ \mathfrak{solution}}}}$

(I) the electrostatic force has a magnitude

$Fe \frac{1}{4 \pi \: e0} • \frac{ {e}^{2} }{ {r}^{2} }$

$= \frac{(9 \times {10}^{9} (N•m²/C²)(1.6× {10}^{ - 19} C²)}{(5.29 \times {10}^{ - 11} )²}$

(2)The magnitude of the gravitational force is

$Fg = G \frac{me \: mp}{ {r}^{2} }$

$\therefore \: \frac{Fe}{Fg} \frac{1}{4 /pi e0}• \frac{e²}{G /: me /: mp}$

$= \frac{(9×10⁹ N•m²/C²}{(6.67× {10}^{ - 11} N•m²/kg²)} \times \\ \frac{(1.6 \times {10}^{ - 19}C)² }{(9.11 \times {10}^{ - 31} kg)(1.67 \times {10}^{ - 27} kg)}$

$= \frac{9 \times 2.56}{6.67 \times 9.11 \times 1.67} \times {10}^{47}$

$= 2.27 \times {10}^{39}$

The magnitude of the electrostatic force is 8.233 × 10-8 N

the ratio of the magnitude of the electrostatic to the gravitational forces is 2.27 × 10³⁹

${ \red{hope \: u \: like \: my \: answer}}$

${ \red{thank \: u \: and \: have \: a \: nyc \: day}}$