Atomic StructureHard
Question
An electron and a proton are accelerated through a potential V. If Pe and Pp are their momentum, then PP: Pe ratio is approximately equal to
Options
A.1: 1836
B.1: 1
C.1836: 1
D.43: 1
Solution
$eV = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m} \Rightarrow p = \sqrt{2meV}$
$\therefore\frac{p_{p}}{p_{e}} = \sqrt{\frac{1836}{1}} = \frac{42.85}{1}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The orbital angular momentum of an electron in 2s orbital is :...The possible set of quantum numbers for which n = 4, l = 3 and s = + 1/2 is...A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency then the number of photons emitted by th...Hyperconjugation involves overlap of the following orbitals...M-2 configuration is 1s2 2s2 2p6 then M+2 configuration will be :-...