- Home
- Standard 11
- Physics
भौतिक स्थिरांकों के निम्नलिखित संयोजन से (अपने साधारण प्रयोग में लिये गये चिन्हों द्वारा प्रदर्शित), केवल वह संयोजन, जो कि इकाइयों के विभित्र निकायों में एक ही मान रखता है
$\frac{{ch}}{{2\pi \varepsilon _0^2}}$
$\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}$
$\frac{{{\mu _0}{\varepsilon _0}G}}{{{c^2}h{e^2}}}$
$\frac{{2\pi \sqrt {{\mu _0}{\varepsilon _0}} h}}{{c{e^2}G}}$
Solution
$\begin{array}{l}
The\,{\rm{Dimensional}}\,{\rm{formulae}}\,{\rm{of}}\\
{\rm{e}}\,{\rm{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} \right]\\
{\varepsilon _0} = \left[ {{M^{ – 1}}{L^{-3}}{T^4}{A^2}} \right]\\
G = \left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]\\
and\,{m_e} = \left[ {{M^1}{L^0}{T^0}} \right]\\
Now,\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}
\end{array}$
$\begin{array}{l}
= \frac{{{{\left[ {{M^0}{L^0}{T^1}{A^1}} \right]}^2}}}{{2\pi \left[ {{M^{ – 1}}{L^{ – 3}}{T^4}{A^2}} \right]\left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]{{\left[ {{M^1}{L^0}{T^o}} \right]}^2}}}\\
= \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^{ – 1 – 1 + 2}}{L^{ – 3 + 3}}{T^{4 – 2}}{A^2}} \right]}}\\
= \frac{{\left[ {{T^2}{A^2}} \right]}}{{2\pi \left[ {{M^0}{L^0}{T^2}{A^2}} \right]}} = \frac{1}{{2\pi }}
\end{array}$
$\begin{array}{l}
\frac{1}{{2\pi }}\,is\,{\rm{Dimensionl}}ess\,thus\,the\,combination\\
\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\end{array}$
would d have the same value in diffierent systems of units