Question

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}

Answer 1

Answer:

Velocity of a proton, $v=1.7\times 10^6\ m/s$    

Explanation:

It is given that,

Potential difference, $V=15\ kV=15\times 10^3\ V$

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 :

$\dfrac{1}{2}mv^2=qV$

q is the charge of proton

m is the mass of proton

$v=\sqrt{\dfrac{2qV}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}$

$v=1695361.75\ m/s$

$v=1.69\times 10^6\ m/s$

or

$v=1.7\times 10^6\ m/s$

So, the velocity of a proton is $1.7\times 10^6\ m/s$. Hence, this is the required solution.