  Question

# In the Bohr's model for the hydrogen-like atoms, how does the orbital radius depend upon the principal quantum numbern ? ZIs the atomic number.    Solution

## The correct option is D The electron of charge e− orbiting a nucleus carrying a positive charge Ze+ will undergo centripetal acceleration due to the electrostatic force of attraction. If the electron's mass is taken as m, and its speed as v, we can write - mv2r=Ze24πϵ0r2 ⇒ r=Ze24πϵ0mv2     ....(1) Here, for different values of v ϵ (0,∞), we can have different values for  r ϵ (0,∞) - but we know from Bohr's postulates that not all values of r are allowed. So there must be some dependence on a discrete variable (a variable that can take only certain values). This discreteness can be obtained by applying Bohr's third postulate - angular momentum can only be integer multiples of h2π. From mechanics, we know that the angular momentum of a point mass m moving in a circle of radius r with a speed v is equal to mvr. We can write - mvr=nh2π ⇒v=nh2πmr        ....(2) Substituting the value of v from equation (2) in equation (1), we obtain the discreteness in the orbital radius that we were looking for – r=Ze24πϵ0mv2=Ze24πϵ0m(nh2πmr)2=Ze24πϵ0m×4π2m2r2h2n2. Cancelling r from both numerators and rearranging, r=h2ϵ0πme2n2Z. The term in the brackets is just a bunch of constants. For Z=1, putting all S.I values of these constants gives us the orbital radius r for n=1 in the hydrogen atom, also called the Bohr radius (denoted by a0), numerically equal to 52.9 pm or 0.529 A ̇.   Suggest corrections  Similar questions  View More...

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