Question

The energy of an electron in the nth orbit of positronium is –

• A. –$\frac{13.6}{n^{2}}$eV
• B. –$\frac{13.6 \times 2}{n^{2}}$eV
• C. –$\frac{13.6 \times 4}{n^{2}}$eV
• D. –$\frac{13.6}{2n^{2}}$eV

Answer 1

Answer: (D)

–$\frac{13.6}{2n^{2}}$eV

Answer 2

Energy E_n of an electron in nth orbit of positronium (a bound system composed of positron and an electron) is given by:

E_n= –$\left( \frac{1}{4\pi\varepsilon_{0}} \right)^{2}$$\left( \frac{mM}{m + M} \right)$

where M = mass of positron

Here, M = m

\E_n= – $\left( \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \right) ^ { 2 }$ $\frac{2\pi^{2}e^{4}}{n^{2}h^{2}}$×$\frac{m}{2}$

=$\frac{(9 \times 10^{9})^{2} \times 2 \times (22/7)^{2} \times (1.6 \times 10^{- 19})^{2} \times (9.1 \times 10^{- 31})}{2n^{2}(6.6 \times 10^{- 34})^{2}}$

= –$\frac{(13.6) \times (1.6 \times 10^{- 19})}{2n^{2}}$joule = –$\frac{13.6}{2n^{2}}$eV