ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2017
  • Hard
  • MCQ

The acceleration of an electron in the first orbit of the hydrogen atom \((z\, = 1)\) is

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}\)

Step-by-step Solution

Detailed explanation

Speed of electron in first orbit \((\mathrm{n}=1)\) of hydrogen atom \((z=1)\),

\(v = \frac{{{e^2}}}{{2{\varepsilon _0}h}}\)

radius of Bohr's first orbit,

\(r=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}} \Rightarrow \varepsilon_{0}=\frac{r \pi m \mathrm{e}^{2}}{h^{2}}\)  ..... \((i)\)

Acceleration of electron,

\(\frac{v^{2}}{r}=\frac{e^{4}}{4 \varepsilon_{0}^{2} h^{2}} \times \frac{\pi m e^{2}}{h^{2} \varepsilon_{0}}\)

\({ = \frac{{{{\text{e}}^4} \times \pi {\text{m}}{{\text{e}}^2}}}{{4{{\text{h}}^4}\varepsilon _0^3}}}\)    ..... \((ii)\)

eliminating \(\varepsilon_{0}\) from eq \((ii)\)

\(=\frac{e^{4} \pi m e^{2} h^{6}}{4 h^{4} r^{3} \pi^{3} m^{3} e^{6}}\) from \(eq^n\) \((i)\)

\(=\frac{h^{2}}{4 \pi^{2} m^{2} r^{3}}\)

Same subject
Explore more questions on app