An electron is revolving in a circular orbit of radius '$r$' in a hydrogen atom. Using Bohr's theory, the angular momentum of the electron is (where '$M$' = magnetic dipole moment,'$m$' = mass of electron,'$e$' = charge of electron).
- A$2mM/e$
- B$mM/e$
- C$e/2mM$
- D$e/mM$
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The radius of the orbit of an electron in a $H$-atom changes from $2.12 \mathring{A}$ to $4.77 \mathring{A}$. The $H$-atom has:
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View SolutionAn electron in the hydrogen atom excites from the $2^{nd}$ orbit to the $4^{th}$ orbit. The change in angular momentum of the electron is (Planck's constant $h = 6.64 \times 10^{-34} \ J \cdot s$).
When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it
Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.
Reason : According to classical physics all moving electrons radiate.
Let $R_1$ be the radius of the second stationary orbit and $R_2$ be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio $\frac{R_1}{R_2}$ is
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