What is the velocity of a proton that has been accelerated by a potential difference of 15 kv? (i)\: \: \: \: \: 9.5\: \times\: 10^5\: m. s^{-1} ( i ) 9.5 × 10 5 m . s − 1 (ii)\: \: \: \: 2.2\: \times\: 10^9\: m. s^{-1} ( i i ) 2.2 × 10 9 m . s − 1 (iii)\: \: \: 3.9\: \times\: 10^8\: m. s^{-1} ( i i i ) 3.9 × 10 8 m . s − 1 (iv)\: \: \: \: 1.7\: \times\: 10^6\: m. s^{-1}

Velocity of a proton,    

Explanation:

It is given that,

Potential difference,

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

q is the charge of proton

m is the mass of proton

or

So, the velocity of a proton is . Hence, this is the required solution.

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explanation: