Atoms & Nuclei

JEE Main level, numerical answer, medium

29 of 239

Question 29 of 239

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd2^{\text{nd}} and 4th4^{\text{th}} orbits of hydrogen atom _______.

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Answer: 32

The magnetic field at the center of a circular orbit is given by B=μ0I2rB = \dfrac{\mu_0 I}{2r}.

The equivalent current due to the revolving electron is
I=eT=ev2πrI = \dfrac{e}{T} = \dfrac{ev}{2\pi r}.

Substituting
II into the expression for BB, we get B=μ0ev4πr2B = \dfrac{\mu_0 ev}{4\pi r^2}.

According to Bohr's model for a hydrogen atom, the velocity of the electron in the
nthn^{\text{th}} orbit is v∝1nv \propto \dfrac{1}{n} and the radius of the orbit is r∝n2r \propto n^2.

Therefore, the magnetic field
BB depends on the principal quantum number nn as B∝1/n(n2)2⇒B∝1n5B \propto \dfrac{1/n}{(n^2)^2} \Rightarrow B \propto \dfrac{1}{n^5}.

The ratio of the magnetic fields in the
2nd2^{\text{nd}} and 4th4^{\text{th}} orbits is B2B4=(n4n2)5=(42)5=25=32\dfrac{B_2}{B_4} = \left(\dfrac{n_4}{n_2}\right)^5 = \left(\dfrac{4}{2}\right)^5 = 2^5 = 32.

Answer:
3232

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