An α - particle accelerated through V volt is fired towards a nucleus. Its distance of closest approach is r. If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

Question

An α - particle accelerated through V volt is fired towards a nucleus. Its distance of closest approach is r. If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be: (A) r (B) 2r (C) r/2 (D) r/4

Tardigrade Admin · Accepted Answer

K.E = P.E ⇒ q1 V = (1/4 π ε0) (q1q2/r) Independent of charge of the particle projected

Q. An $α$ - particle accelerated through $V$ volt is fired towards a nucleus. Its distance of closest approach is $r$ . If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

r

2r

r/2

r/4

K.E = P.E $\Rightarrow$ $q_{1} V = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r}$
Independent of charge of the particle projected

Q. An α - particle accelerated through V volt is fired towards a nucleus. Its distance of closest approach is r. If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

r

2r

r/2

r/4

K.E = P.E ⇒ q1​V=4πε0​1​rq1​q2​​ Independent of charge of the particle projected

Q. An $α$ - particle accelerated through $V$ volt is fired towards a nucleus. Its distance of closest approach is $r$ . If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

K.E = P.E $\Rightarrow$ $q_{1} V = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r}$
Independent of charge of the particle projected