A) X-rays
B) Infrared rays
C) Ultravioiet rays
D) \[\gamma \]-rays
Correct Answer: A
Solution :
Given, energy of EM waves is of the order of i.e. |
\[E=hv=h\times \frac{c}{\lambda }\] |
\[\Rightarrow \] \[\lambda =\frac{h\times c}{E}=\frac{6.624\times {{10}^{-34}}\times 3\times {{10}^{18}}}{15\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}\] |
\[=\frac{1.3248\times {{10}^{-29}}}{1.6\times {{10}^{-19}}}={{0.82810}^{-10}}m\] |
\[=0.828\overset{{}^\circ }{\mathop{\text{A}}}\,\] \[[1\overset{{}^\circ }{\mathop{A}}\,={{10}^{-10}}m]\] |
\[\lambda =0.828\overset{{}^\circ }{\mathop{A}}\,\] |
Thus, this spectrum is a part of X-rays. |
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