Question

a proton is revolving on a circular path radius 0.1m if the value of centripetal force working on it is 40*10^-13 Newtown then find speed of proton and rotational frequency. mass of proton is 1.6*10^-27​

Answer 1

Answer

Velocity of proton = 1.58 × 10⁷ m/s

Explanation

Given:-

• Mass of proton, m = 1.6 × 10⁻²⁷ kg
• Radius of the circular path, r = 0.1 m
• Centripetal force on it, F = 40 × 10⁻¹³ N

To find:-

• Velocity of the proton, v =?

Formula required:-

• Formula for centripetal force

        F = mv²/r

[ Where F is centripetal force, m is mass of body, v is velocity and r is radius of circular path ]

Solution:-

Using the formula for centripetal force

→ F = mv²/r

→ 40 × 10⁻¹³ = (1.6 × 10⁻²⁷) × v² / 0.1

→ 40 × 10⁻¹³ = 1.6 × 10⁻²⁶ × v²

→ v² = ( 40 × 10⁻¹³ ) / ( 1.6 × 10⁻²⁶ )

→ v² = 25 × 10¹³ = 2.5 × 10¹⁴

→ v = 1.58 × 10⁷ m/s

Therefore,

• Velocity of the proton is 1.58 × 10⁷ m/s. (approx.)

Answer 2

Formulas:

v = rw

F = (mv²)/r

Working:

a) Speed of proton (v)

$\frac{f \times r}{m} = {v}^{2} \\ v = \sqrt{ \frac{(40 \times {10}^{ - 13}) \times 0.1 }{1.6 \times {10}^{ - 27} } } \\ v = 1.58 \times {10}^{7} m/s$

b) rotational frequency (w)

$w = \frac{v}{r} \\ w = \frac{1.58 \times {10}^{7} }{0.1} \\ w = 1.58 \times {10}^{8}$