JEE Main Structure of Atom Practice Questions With Solutions

Match List I with List II

Choose the correct answer from the options given below:

The correct set of four quantum numbers for the valence electron of rubidium atom $(\mathrm{Z}=37)$\left(\right. Z = 37 \left.\right) is :

The energy of an electron in the first Bohr orbit of hydrogen atom is $-2.18\times {10}^{-18}\mathrm{J}$- 2.18 \times 10^{- 18} \textrm{ } J. Its energy in the third Bohr orbit is ____________.

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :

Assertion A : In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface.

Reason R : When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place.

In the light of the above statements, choose the most appropriate answer from the options given below :

Henry Moseley studied characteristic X-ray spectra of elements. The graph which represents his observation correctly is

Given $v=$v = frequency of $\mathrm{X}$X-ray emitted

Z = atomic number

If the radius of the first orbit of hydrogen atom is ${\alpha }_0$\alpha_{0}, then de Broglie's wavelength of electron in $3^{\text{rd }}$3^{\text{rd}\textrm{ }} orbit is :

Which one of the following sets of ions represents a collection of isoelectronic species?

(Given : Atomic Number : $\mathrm{F}:9,\mathrm{Cl}:17,\mathrm{Na}=11,\mathrm{Mg}=12,\mathrm{Al}=13,\mathrm{K}=19,\mathrm{Ca}=20,\mathrm{Sc}=21$F : 9 , Cl : 17 , Na = 11 , Mg = 12 , Al = 13 , K = 19 , Ca = 20 , Sc = 21)

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $\mathrm{n}=4$n = 4 to $\mathrm{n}=2$n = 2 of ${\mathrm{He}}^{+}$He^{+} spectrum