Atomic StructureHard
Question
Two electrons are moving in orbits of two hydrogen like atoms with speeds $3 \times 10^{5}\text{ }m/s$ and $2.5 \times 10^{5}\text{ }m/s$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are $\_\_\_\_$ respectively.
Options
A.6 and 5
B.9 and 8
C.8 and 10
D.10 and 12
Solution
(4) 10 and 12
Sol. $V \propto \frac{Z}{n}$
$$r \propto \frac{n^{2}}{Z} $$Thus ; $r \propto \frac{n}{V}$
Radii are same then
$${\frac{n_{1}}{{\text{ }V}_{1}} = \frac{n_{2}}{{\text{ }V}_{2}} }{\frac{n_{1}}{n_{2}} = \frac{3 \times 10^{5}}{2.5 \times 10^{5}} = \frac{6}{5} }$$Possible order is 6 and 5
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
he decay constant of radioactive sample is λ. The half-life and mean-life of the sample are respectively given by...Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is :...A light emitting diode (LED) has a voltage drop of 2 volt across it and passes a current of 10 mA.When it operates with ...The maximum kinetic energy of photoelectrons emitted from a surface is $4 \text{ eV}$ when photons of energy $6 \text{ e...In photoelectric effected, the electrons are ejected from metals if the incident light has a certain minimum...