Physics · Atoms · NEET
The radius is proportional to n squared, NOT to n. From the derivation r_n = 0.53 n^2 / Z angstrom, so as you go from n=1 to n=2 to n=3 the radii are in ratio 1 : 4 : 9 (the squares of 1, 2, 3). Students often lose marks by writing r proportional to n. Remember: radius scales as n^2, speed scales as 1/n, and energy scales as 1/n^2.
A larger nuclear charge Ze pulls the electron in more strongly, so the orbit shrinks. In the full formula r_n = (n^2 h^2 epsilon_0)/(pi m Z e^2), Z sits in the denominator, giving r_n = 0.53 n^2 / Z angstrom. For hydrogen Z=1 so it disappears. For He+ (Z=2) every orbit is half the hydrogen size; for Li2+ (Z=3) it is one-third.
The Bohr radius a_0 is the radius of the first orbit (n=1) of hydrogen (Z=1): a_0 = 0.53 angstrom = 0.53 x 10^-10 m = 5.3 x 10^-11 m = 53 pm. NCERT and NEET accept 0.53 A or 0.51 A depending on rounding of constants. This is the smallest allowed orbit, the ground state.
Yes. The mass m sits in the denominator, so r_n is proportional to 1/m. If you replace the electron with a heavier particle like a muon (mass = 207 times the electron mass), the orbit shrinks by 207 times. This is exactly the NEET 2019 muon question: r_muon = 0.51 A / 207 = 2.56 x 10^-13 m.
You combine two physical facts. (1) Coulomb force = centripetal force: (1/4 pi epsilon_0)(Ze^2/r^2) = mv^2/r keeps the electron in a circle. (2) Bohr's quantisation: mvr = nh/2pi restricts the angular momentum to whole-number multiples of h/2pi. Solving these two together eliminates v and gives r_n proportional to n^2.
The radius of innermost orbit of hydrogen atom is 5.3 x 10^-11 m. What is the radius of the third allowed orbit of hydrogen atom?
The radius of the first permitted Bohr orbit for the electron in a hydrogen atom is 0.51 A and its ground state energy is -13.6 eV. If the electron in hydrogen is replaced by a muon (charge same as electron, mass 207 m_e), the first Bohr radius will be:
An electron is revolving in an excited state of a hydrogen atom with velocity sqrt(25.6) x 10^5 m/s. The radius of the orbit is x x 10^-9 m. The value of x is: [m_e = 9 x 10^-31 kg, e = 1.6 x 10^-19 C, 1/(4 pi epsilon_0) = 9 x 10^9 N m^2 C^-2]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
r_n = (n^2 h^2 epsilon_0) / (pi m Z e^2), which for hydrogen (Z=1) simplifies to r_n = 0.53 n^2 angstrom. In general r_n = 0.53 n^2 / Z angstrom.
The Bohr radius is the first orbit of hydrogen: a_0 = 0.53 angstrom = 5.3 x 10^-11 m = 53 pm. It is the smallest allowed electron orbit.
r_n is proportional to n^2 (radius grows as the square of the orbit number), proportional to 1/Z (higher nuclear charge shrinks the orbit) and proportional to 1/m (heavier particle means smaller orbit).
r_2 = 0.53 x 2^2 = 0.53 x 4 = 2.12 angstrom = 2.12 x 10^-10 m. That is 4 times the first orbit.
Bohr postulated special stationary orbits where the electron does not radiate energy. The quantisation condition mvr = nh/2pi fixes discrete allowed radii, the smallest being the Bohr radius, so the electron cannot spiral inward.
