Atoms & Nuclei

JEE Main level, single correct, medium

31 of 239

Question 31 of 239

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3×105 m/s3 \times 10^{5} \mathrm{~m} / \mathrm{s} and 2.5×105 m/s2.5 \times 10^{5} \mathrm{~m} / \mathrm{s} respectively. If the radii of these orbits are nearly same then the possible order of energy states are ____\_\_\_\_ respectively.

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Answer: (D) 6 and 5

For hydrogen-like atoms:
V∝ZnV \propto \frac{Z}{n} and r∝n2Zr \propto \frac{n^2}{Z}
Thus r∝nVr \propto \frac{n}{V}
Since the radii are nearly the same:
n1V1=n2V2\frac{n_1}{V_1} = \frac{n_2}{V_2}
n1n2=V1V2=3×1052.5×105=65\frac{n_1}{n_2} = \frac{V_1}{V_2} = \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{6}{5}
Possible order of energy states is 6 and 5.

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