An α - particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α - particle to the nucleus will be

Question

An α - particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α - particle to the nucleus will be (A) 14.4(Z/V) overset° A (B) 14.4(Z/V) m (C) 14.4(Z/V) cm (D) all of these

Tardigrade Admin · Accepted Answer

K.E. = P.E. = qV ⇒ 2eV=(K (Z e) (2 e)/d)=qV ∴ d=(9 × 1 09 × Z × e × 2 e/2 e V) ∴ d=(9 × 1 09 × 1.6 × 1 0- 19 × Z/V) d=14.4× 10- 10((Z/V))m

Q. An $α -$ particle after passing through a potential difference of $V$ volts collides with a nucleus. If the atomic number of the nucleus is $Z$ then the distance of closest approach of $α -$ particle to the nucleus will be

$14.4 \frac{Z}{V} A^{\circ}$

$14.4 \frac{Z}{V} m$

$14.4 \frac{Z}{V} c m$

all of these

K.E. = P.E. = qV
$\Rightarrow 2 e V = \frac{K ( Z e ) ( 2 e )}{d} = q V$
$∴ d = \frac{9 \times 1 0 ^{9} \times Z \times e \times 2 e}{2 e V}$
$∴ d = \frac{9 \times 1 0 ^{9} \times 1.6 \times 1 0 ^{- 19} \times Z}{V}$
$d = 14.4 \times 1 0^{- 10} (\frac{Z}{V}) m$

Q. An α− particle after passing through a potential difference of V volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α− particle to the nucleus will be

14.4VZ​A∘

14.4VZ​m

14.4VZ​cm