Two electrons are moving in orbits of two hydrogen-like atoms with speeds and respectively. If the radii of these orbits are nearly same, then the possible order of energy states are _______ respectively. [2026]

Question

Two electrons are moving in orbits of two hydrogen-like atoms with speeds $3 \times 10^{5} m/s$ and $2.5 \times 10^{5} m/s$ respectively. If the radii of these orbits are nearly same, then the possible order of energy states are _______ respectively. [2026]

Smart Achievers · Accepted Answer

(2)

$V \propto \frac{Z}{n}$

$r \propto \frac{n^{2}}{Z}$

$Thus, r \propto \frac{n}{V}$

$Radii are same then$

$\frac{n_{1}}{V_{1}} = \frac{n_{2}}{V_{2}}$

$\frac{n_{1}}{n_{2}} = \frac{3 \times 10^{5}}{2.5 \times 10^{5}} = \frac{6}{5}$

$Possible order is 6 and 5$

Solution

Q.
Two electrons are moving in orbits of two hydrogen-like atoms with speeds $3 \times 10^{5} m/s$ and $2.5 \times 10^{5} m/s$ respectively. If the radii of these orbits are nearly same, then the possible order of energy states are _______ respectively. [2026]

1 8 and 10

2 6 and 5

3 10 and 12

4 9 and 8

Ans.
(2)

$V \propto \frac{Z}{n}$

$r \propto \frac{n^{2}}{Z}$

$Thus, r \propto \frac{n}{V}$

$Radii are same then$

$\frac{n_{1}}{V_{1}} = \frac{n_{2}}{V_{2}}$

$\frac{n_{1}}{n_{2}} = \frac{3 \times 10^{5}}{2.5 \times 10^{5}} = \frac{6}{5}$

$Possible order is 6 and 5$

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