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 Niels Bohr, this electron moves in a stationary orbit, it emits no electromagnetic radiation. The angular momentum of the electron is quantized, i.e., $mvr = (nh / 2 π)$ , where m=mass of the electron, v = velocity of the electron in the orbit, r = radius of the orbit, and n=1,2,3...when transition takes place from $K^{t h}$ orbit to $J^{th}$ orbit, energy photon is emitted. If the wave length of the emitted photon is $λ,$ we find that $\frac{1}{λ} = R [\frac{1}{J^{2}} - \frac{1}{K^{2}}]$ , where 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 $P = 2 n (h/2π)$ , i.e. an even multiple of $(h / 2 π) .$

Question

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 Niels Bohr, this electron moves in a stationary orbit, it emits no electromagnetic radiation. The angular momentum of the electron is quantized, i.e., $mvr = (nh / 2 π)$ , where m=mass of the electron, v = velocity of the electron in the orbit, r = radius of the orbit, and n=1,2,3...when transition takes place from $K^{t h}$ orbit to $J^{th}$ orbit, energy photon is emitted. If the wave length of the emitted photon is $λ,$ we find that $\frac{1}{λ} = R [\frac{1}{J^{2}} - \frac{1}{K^{2}}]$ , where 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 $P = 2 n (h/2π)$ , i.e. an even multiple of $(h / 2 π) .$

Infinity Learn · Accepted Answer

In our world, the ionization potential energy of a hydrogen atom is $13.6 eV$ . On the other planet, this ionization potential energy will be $3.4 eV$

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The minimum permissible radius of the orbit will be $\frac{4 ε_{0} h^{2}}{m π e^{2}}$

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In our world, the velocity electron is $v_{0}$ when the hydrogen atom is in the ground state. The velocity of electron in this state on the other planet should be $v_{0} / 2$

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