You visited us 1 times! Enjoying our articles? Unlock Full Access!

Characteristics of Particles of Matter

State the uni...

Question

State the universal law of gravitation. The mass of the electron and proton in a Hydrogen atom is given by 9×10^-31 Kg and 1.9×10^-27 Kg respectively which is separated by a distance 6×10^-11m. Calculate the gravitational force of attraction between them? [3 MARKS]

Open in App

Solution

Definition: 1 Mark
Problem: 2 Mark

1.The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

2. The given parameters are:
M_e=9×10^-31 Kg and M_p=1.9×10^-27 Kg, G=6.673×10^-11Nm²/Kg², r=6×10^-11m
$F = \frac{G m_{1} m_{2}}{r^{2}}$

F=6.673×10−11×9×10−31×1.9×10−27(6×10−11)2

F=3.16×10^-47N

Suggest Corrections

Similar questions

5 \times 10^{- 11} m

= 1.6 \times 10^{- 19} C

= 9.1 \times 10^{- 31}

= 1.6 \times 10^{- 27} k g, G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})

5.3 \times 10^{- 11} m

= 9.1 \times 10^{- 31} k g, m p = 1.67 \times 10^{- 27} k g, e = 1.6 \times 10^{- 19} C

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

5 \times 10^{- 11} m

= 1.6 \times 10^{- 19} C

= 9.1 \times 10^{- 31}

= 1.6 \times 10^{- 27} k g, G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})

m_{p} = 1.67 \times 10^{- 27} k g

m_{e} = 9.1 \times 10^{- 31} k g

q_{p} = 1.6 \times 10^{- 19} C

9.1 \times 10^{- 31}

10 A^{0}

1 A^{0} = 10^{- 10} m .

Join BYJU'S Learning Program