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JEE Advance 2025 Paper 1

JEE Advance · 2025

94 questions · One at a time

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SHORT ANSWER4 marks

Electron in n=3n = 3   orbit has same de Broglie wavelength as a thermal neutron at TT  . T=Z2h2απ2a02mNkBT = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}  . Find α\alpha  .

Answer

72
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