An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

<p>When an electron is projected perpendicular to a uniform magnetic field $ B $, it travels in a circular path. According to Bohr's quantization rule, the radius of the electron's orbit in the first excited state can be determined using the following approach:</p> <p><p>The radius $ r $ in a magnetic field is expressed as:</p> <p>$ r = \frac{mv}{eB} $</p> <p>where $ m $ is the mass of the electron, $ v $ is its velocity, $ e $ is its charge, and $ B $ is the magnetic field strength.</p></p> <p><p>According to Bohr's quantization condition:</p> <p>$ mvr = \frac{nh}{2\pi} $</p> <p>Substituting the expression for $ r $:</p> <p>$ (eBr)r = \frac{nh}{2\pi} $</p> <p>This simplifies to the expression for the radius:</p> <p>$ r = \sqrt{\frac{nh}{2\pi eB}} $</p></p> <p><p>In the first excited state, the principal quantum number $ n $ is 2. So, the radius becomes:</p> <p>$ r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}} $ </p></p> <p>Thus, for the first excited state, the radius of the electron's orbit is $ \sqrt{\frac{h}{\pi eB}} $.</p>