Atomic StructureHard
Question
Two particles A and B having same e/m ratio are projected towards silver nucleus in different experiments with the same speed. The distance of closest approach will be
Options
A.same for both.
B.greater for A.
C.greater for B.
D.depends on speed.
Solution
$r = \frac{K.q_{1}q_{2}}{\left( \frac{1}{2}mv^{2} \right)}$
As $\left( \frac{q_{1}}{m} \right)$, q2 and v, all are same, r is same.
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The orbital angular momentum of an electron is 2s orbital is...An amount of 1.75 × 10–4 mole of HI decomposes by the absorption of photons of wavelength 2500 Å. If one molecule is dec...If n and l are respectively the principal and azimuthal quantum numbers, then the expression for calculating the total n...The work functions of two metals ( $M_{A}$ and $M_{B}$ ) are in the $1:2$ ratio. When these metals are exposed to photon...Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain metal (work function = 2.0...