In Bohr model of hydrogen atom

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the radius of nth orbit is proportional to n2

the total energy of electron in nth orbit is proportional to n

the angular momentum of the electron in an orbit is an integral multiple of h /2π

the magnitude of the potential energy of an electron in any orbit is greater than its kinetic energy

answer is A.

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Detailed Solution

rn=n2h2ε0πme2 rn∝n2 mvr=nh/2π Angular momentum is integral multiple of h/2π) K.E. =me48n2h2ε02 and P.E. =−me48n2h2ε02 ∣ P.E. |>| K.E. ∣

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