Radius of a certain orbit of hydrogen atom is 8.48 Å. If energy of electron in this orbit is E / x , then x = \ \ \ \ \ \ \ \ . (Given a 0 =0.529overset 0 mathop A \,,\,\,\,E= energy of electron in ground state)

Physics · Atomic Physics

JEE Main 2024 — 6 April, Shift 1 — Question 56

Radius of a certain orbit of hydrogen atom is 8.48 Å. If energy of electron in this orbit is $\mathrm{E} / \mathrm{x}$ , then $\mathrm{x}=$ $\_\_\_\_\_\_\_\_$ . (Given ${{a}_{0}}=0.529\overset{0}{\mathop{A}}\,,\,\,\,E=$ energy of electron in ground state)

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

We know

$\mathrm{r}=0.529 \frac{\mathrm{n}^{2}}{\mathrm{Z}} \Rightarrow 8.48=0.529 \frac{\mathrm{n}^{2}}{1}$ $\mathrm{n}^{2}=16 \Rightarrow \mathrm{n}=4$

We know

$\mathrm{E} \propto \frac{1}{\mathrm{n}^{2}}$

$\mathrm{E}_{\mathrm{n}^{\mathrm{th}}}=\frac{\mathrm{E}}{16}$

$\mathrm{x}=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Atomic Physics
Topic: Rutherford's and Bohr's Model of Atom

Answer: 16

Step-by-step solution

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