Board Boosters

The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

100 questions · clear filters

PhysicsClass 122 marksmedium

Atoms

State the main conclusions of Rutherford's alpha-particle scattering experiment.

Reveal model answer + marking points▾

From the scattering of alpha particles by a thin gold foil, Rutherford concluded that: (1) most of the atom is empty space, since most alpha particles passed straight through; (2) the entire positive charge and almost all the mass of the atom are concentrated in a very small central region called the nucleus, since a few alpha particles were deflected through large angles; and (3) the electrons revolve around the nucleus, and the size of the nucleus is very small compared with the size of the atom.

Marking-scheme points

  • ✓Most of the atom is empty space
  • ✓Positive charge and mass concentrated in a tiny nucleus
  • ✓Electrons revolve around the nucleus
Still unsure? Ask the AI tutor →
PhysicsClass 123 marksmedium

Atoms

State the postulates of Bohr's model of the hydrogen atom.

Reveal model answer + marking points▾

Bohr's postulates are: (1) The electron revolves around the nucleus only in certain fixed circular orbits called stationary states, in which it does not radiate energy. (2) Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/(2 pi), that is, m v r = n h/(2 pi) (quantisation of angular momentum). (3) Energy is emitted or absorbed only when the electron jumps from one orbit to another, the energy of the emitted or absorbed photon being h f = E2 - E1.

m v r = n h/(2 pi)

Marking-scheme points

  • ✓Electrons revolve in fixed stationary orbits without radiating
  • ✓Angular momentum quantised: m v r = n h/(2 pi)
  • ✓Energy change on jump: h f = E2 - E1
Still unsure? Ask the AI tutor →
PhysicsClass 122 marksmedium

Atoms

The energy of an electron in the ground state of hydrogen is -13.6 eV. Calculate the energy of the electron in the second orbit (n = 2).

Reveal model answer + marking points▾

The energy of the electron in the nth orbit of hydrogen is En = -13.6/n^2 eV. For n = 2, E2 = -13.6/2^2 = -13.6/4 = -3.4 eV. The negative sign shows that the electron is bound to the nucleus, and the energy increases (becomes less negative) as n increases.

En = -13.6/n^2 eV

Marking-scheme points

  • ✓En = -13.6/n^2 eV
  • ✓E2 = -13.6/4
  • ✓E2 = -3.4 eV
Still unsure? Ask the AI tutor →
← PrevPage 5 of 5

You are more ready than you feel.

One question at a time is how every topper started. Bookmark this, revise a few each day, and watch the fear shrink. And if a friend is stressing about boards — send this their way. You both win.