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NEET UG PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Application +4 | -1
The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of 4.14×10−144.14 \times {10^{ - 14}} J is (Boltzmann constant
=1.38×10−23  JK−1)  \times {10^{ - 23}}\;\mathrm{J}{\mathrm{K}^{ - 1}})\mathrm{\;}
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Correct option: A
2×1092 \times {10^9} K
Option A (2×1092 \times {10^9} K) is correct because it is calculated using the formula E=32kTE = \frac{3}{2}kT.

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