SATHEE: Atoms And Nuclei Question 71

Atoms And Nuclei Question 71

Question: The radius of hydrogen atom in its ground state is $5.3 \times 10^{- 11} m$ . After collision with an electron it is found to have a radius of $21.2 \times 10^{- 11} m$ . What is the principal quantum number n of the final state of the atom [CBSE PMT 1994; CPMT 2001; MH CET 2000]

Options:

A) n = 4

B) n = 2

C) n = 16

D) n = 3

Show Answer

Answer:

Correct Answer: B

Solution:

$r \propto n^{2} i . e . \frac{r_{f}}{r_{i}} = {(\frac{n_{f}}{n_{i}})}^{2}$
Þ $\frac{21.2 \times 10^{- 11}}{5.3 \times 10^{- 11}} = {(\frac{n}{1})}^{2}$
Þ $n^{2} = 4$
Þ n = 2

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Atoms And Nuclei Question 71

Question: The radius of hydrogen atom in its ground state is 5.3×10−11m . After collision with an electron it is found to have a radius of 21.2×10−11m . What is the principal quantum number n of the final state of the atom [CBSE PMT 1994; CPMT 2001; MH CET 2000]

Options:

Answer:

Solution:

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Question: The radius of hydrogen atom in its ground state is $5.3 \times 10^{- 11} m$ . After collision with an electron it is found to have a radius of $21.2 \times 10^{- 11} m$ . What is the principal quantum number n of the final state of the atom [CBSE PMT 1994; CPMT 2001; MH CET 2000]