b
દ બ્રોગ્લી તરંગ લંબાઈ \(\lambda \,\, = \,\, \frac{h}{{\sqrt {3m{k_B}T} }}\)

અહીં,\( m\,\, = \,\,9.11\,\, \times \,\,1{0^{ - 31}}kg,\) \({k_B} = \,\,1.38\,\, \times \,\,{10^{ - 23}}\,J/K\) અને \(T\,\, = \,\,27\,\, + \,\,273\,\, = \,\,300\,\,K\)

આથી, \(\lambda \,\,\, = \,\,\,\,\frac{{6.63\,\, \times \,\,{{10}^{ - 34}}}}{{\sqrt {3\,\,(9.11\,\, \times \,\,{{10}^{ - 31}})\,\,(1.38\,\, \times \,\,{{10}^{ - 23}})\,\,300} }}\,\)

\(m\,\, = \,\,6.2\,\, \times \,\,1{0^{ - 9}}m\)

બેઇલેક્ટ્રોન વચ્ચેનું સરેરશ અંતર ,\(\,r\,\, = \,\,2\,\, \times \,1{0^{ - 10}}\,m\,;\,\,\,r\,\, = \,\,2\,\, \times \,\,{10^{ - 10}}m\,\,;\,\,\)

\(\,\frac{r}{{\lambda \,}}\,\, = \,\,\,\frac{{2\,\, \times \,\,{{10}^{ - 10}}m}}{{6.2\,\, \times \,\,{{10}^{ - 9}}m}}\,\, = \,\,0.03\)