Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niel Bohr, this electron moves in a stationary orbit. When this electron is in the stationary orbit, it emits no electromagnetic radiation. The angular momentum of the electron is quantized, i.e., \( m v r=n\left(\frac{h}{2 \pi}\right) \), where, \( \mathrm{m}= \) mass of the electron, \( \mathrm{v}= \) velocity of the electron in the orbit, \( r= \) radius of the orbit, \( \mathrm{n}=1,2,3 \ldots \). When transition takes placed from Kth orbit to Jth orbit, energy photon is emitted. If the wavelength of the emitted photon is \( \lambda \), we find that \( \frac{1}{\lambda}=R\left(\frac{1}{J^{2}}-\frac{1}{K^{2}}\right) \) where \( \mathrm{R} \) is Rydberg's constant. On a different planet, the hydrogen atom's structure was somewhat different from ours. There the angular momentum of electron was \( P=2 n\left(\frac{h}{2 \pi}\right) \), i.e., an even multiple of \( \left(\frac{h}{2 \pi}\right) \).
Answer the following questions based on above passage:
The minimum permissible radius of the orbit will be
(1) \( \frac{2 \varepsilon_{0} h^{2}}{m \pi e^{2}} \)
(2) \( \frac{4 \varepsilon_{0} h^{2}}{m \pi e^{2}} \)
(3) \( \frac{\varepsilon_{0} h^{2}}{m \pi e^{2}} \)
(4) \( \frac{\varepsilon_{0} h^{2}}{2 m \pi e^{2}} \)
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