The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is $...... \text{ } \mathring{A}$ (radius of the first orbit of hydrogen atom $= 0.53 \text{ } \mathring{A}$).

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:

According to Bohr's atomic model,the angular momentum of an electron in the $5^{th}$ orbit of an $H$-atom is:

$A$ hydrogen atom,initially at rest in its ground state,absorbs a photon of frequency $v_1$ and ejects the electron with a kinetic energy of $10 eV$. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $v_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of $5 eV$. It is given that the positron has the same mass as that of the electron and the positronium atom can be considered as a Bohr atom,in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process,the difference between the two photon energies (in $eV$) is $....$ (in $.80$)

In a hydrogen atom,an electron of mass $9.1 \times 10^{-31} \ kg$ revolves about a proton in a circular orbit of radius $0.53 \ \mathring{A}$. The radial acceleration and angular velocity of the electron are respectively:

The figure shows standing de-Broglie waves due to the revolution of an electron in a certain orbit of a hydrogen atom. Then,the expression for the orbit radius is (All notations have their usual meanings).