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Q.

The radius of the electron orbit in the ground state of hydrogen is 0.53 Å. Due to collision with a neutron, the electron shifts to a state in which its orbital radius becomes 4.77 Å . The principal quantum number of the final state of hydrogen is:

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a

$n = 1$

b

$n = 3$

c

$n = 4$

d

$n = 2$

answer is C.

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Detailed Solution

$r_{n} \propto n^{2}$ .

Therefore $n \propto \sqrt{r_{n}} .$

If $r_{i}$ and $r_{f}$ are the radii of the initial and final states, then:

$n = \sqrt{\frac{r_{f}}{r_{i}}} = \sqrt{\frac{4.77}{0.53}} = \sqrt{9} = 3$

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The radius of the electron orbit in the ground state of hydrogen is 0.53 Å. Due to collision with a neutron, the electron shifts to a state in which its orbital radius becomes 4.77 Å . The principal quantum number of the final state of hydrogen is: