Daily Chemistry Problems for JEE - Day 5

Chemistry Level 2

The potential energy of an electron in the 2nd Bohr's orbit of a hydrogen atom is :

Details: a 0 a_0 is Bohr's radius, h is the Planck's constant and m is the mass of an electron.

− h 2 2 π 2 m a 0 2 \frac{-h^2}{2{\pi}^2m{a_0}^2} − h 2 8 π 2 m a 0 2 \frac{-h^2}{8{\pi}^2m{a_0}^2} − h 2 16 π 2 m a 0 2 \frac{-h^2}{16{\pi}^2m{a_0}^2} − h 2 32 π 2 m a 0 2 \frac{-h^2}{32{\pi}^2m{a_0}^2}

Sahil Bansal by Sahil Bansal , 21, India

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