Given maximum are temperature on the order of \(\displaystyle{9}\cdot{3}\times{10}^{{{3}}}{K}\)

Nuclear explosion produces on temperature order of \(\displaystyle{9}\cdot{2}\times{10}^{{{6}}}{K}\)

a)We know that

\(\displaystyle{X}_{{\max}}=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}{K}}}{{T}}\)

\(\displaystyle=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}\cdot{K}}}{{{9}\cdot{3}\times{10}^{{{3}}}}}\)

\(\displaystyle={3}\cdot{116}\times{10}^{{-{7}}}{m}\)

\(\displaystyle={311}\cdot{6}\times{10}^{{-{9}}}{m}={311}\cdot{6}{n}{m}\)

b) \(\displaystyle\lambda_{{\max}}=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}\cdot{K}}}{{T}}\)

\(\displaystyle=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}}}{{{9.2}\times{10}^{{{6}}}}}\)

\(\displaystyle={3}\cdot{15}\times{10}^{{-{10}}}{m}\)

\(\displaystyle={315}\cdot{0}\times{10}^{{-{12}}}{m}={315}\cdot{0}\pm\)

Nuclear explosion produces on temperature order of \(\displaystyle{9}\cdot{2}\times{10}^{{{6}}}{K}\)

a)We know that

\(\displaystyle{X}_{{\max}}=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}{K}}}{{T}}\)

\(\displaystyle=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}\cdot{K}}}{{{9}\cdot{3}\times{10}^{{{3}}}}}\)

\(\displaystyle={3}\cdot{116}\times{10}^{{-{7}}}{m}\)

\(\displaystyle={311}\cdot{6}\times{10}^{{-{9}}}{m}={311}\cdot{6}{n}{m}\)

b) \(\displaystyle\lambda_{{\max}}=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}{m}\cdot{K}}}{{T}}\)

\(\displaystyle=\frac{{{0}\cdot{2898}\times{10}^{{-{2}}}}}{{{9.2}\times{10}^{{{6}}}}}\)

\(\displaystyle={3}\cdot{15}\times{10}^{{-{10}}}{m}\)

\(\displaystyle={315}\cdot{0}\times{10}^{{-{12}}}{m}={315}\cdot{0}\pm\)