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 Multiple Choice QuestionsMultiple Choice Questions

1.

Continuous emission spectrum, emitted wavelength about spectrum, the range depends upon:

  • temperature of the source

  • material of the source

  • surface area of the source

  • all of the above


A.

temperature of the source


2.

The radius of electron second stationary orbit in Bohr's atom is R. The radius of third orbit will be

  • 2.25R

  • 3R

  • R/3

  • 9R


D.

9R

rn=n2mh2π24πεoe2rnn2For second orbit n=2r2= Rfor third orbit n=3r3=32Rr3=9R


3.

Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be

  • two 

  • three

  • four

  • one


B.

three

Ionization energy corresponding to ionization potential  = -13.6 eV.
Photon energy incident = 12.1 eV
So, the energy of electron in excited state
    = -13.6 + 12.1 = -1.5 eV
i.e.,      straight E subscript straight n space equals space minus fraction numerator 13.6 over denominator straight n squared end fraction space rightwards double arrow space space straight n squared space equals space minus fraction numerator 13.6 over denominator negative 1.5 end fraction space equals space 9
therefore space space space space space space straight n space equals space 3
i.e., energy of electron in excited state corresponds to third orbit.
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.

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4.

The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the electron in the state?

  • 0 eV

  • -27.2 eV

  • 1 eV

  • 2 eV


B.

-27.2 eV

Total energy = kinetic energy + potential energy

Potential energy = 2 total energy

                        = -2 × 13.6

                        = -27.2 eV


5.

The electric potential at the surface of an atomic nucleus (Z = 50) of radius 9.0×10-15m is

  • 8×106 V

  • 80V

  • 9×105 V

  • 9V


A.

8×106 V

V=14πεoQR=14πεoZeR                    =9×109×50×1.6×10-199.0×10-15                  V  =8×106Volt


6.

Energy required to break one bond in DNA is approximately

  • ∼ 1 eV

  • ∼ 0.1 eV

  • ∼ 0.01 eV

  • ∼ 2.1 eV


A.

∼ 1 eV

Energy required to break one bond in DNA is ∼ 1 eV


7.

Radius of first Bohr orbit is r. What is the radius of 2nd Bohr orbit ?

  • 8 r

  • 2 r

  • 4 r

  • 22 r


C.

4 r

Radius of Bohr orbit 

vn = n2mh24πεo1h2π

r ∝ n2

r1r2 = n1n22 r1r2=122rr2 =  14r2 = 4r


8.

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.

  • if both assertion and reason are true and the reason is the correct explanation of the assertion

  • if both assertion and reason are true and the reason is not the correct explanation of the assertion

  • if assertion is true but reason is false

  • if both assertion and reason are false statements


B.

if both assertion and reason are true and the reason is not the correct explanation of the assertion

According to classical physics, all moving charged particle radiate electromagnetic radiation. So moving electrons will also radiate energy. If we see the atomic structure we find that electrons revolve around the nucleus in some particular orbits. Bohr termed these orbits as the stationary orbits as the electrons do not radiate energy as long as they are moving in these orbits. This is one of Bohr's postulates. This postulate is based on the fact that if the moving electrons radiate thereby losing energy, they have got a chance to finally fall back onto the nucleus and the atom will be collapsed.


9.

The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10-34 Js, me =9.1×10-31 kg)

  • 5.47Ao

  • 10.9Ao

  • 2.7Ao

  • None of these


A.

5.47Ao

Here:- E =5ev=5×1.6×10-19Jλ=h2mE  =6.6×10-342×9.1×10-31×5×1.6×10-19=6.6×10-341.2066×10-24=5.47×10-10 m=5.47Ao


10.

If the temperature of a black body is doubled, the wavelength at which the spectral radiancy has its maximum, is 

  • doubled

  • halved

  • quadrupled

  • unchanged


B.

halved

From Wein's displacement law,

        λmT = constant 

    λmT  =  λ'mT

    λ'm  =  λmTT'

    λ'm  =  λmT2T

    λ'm  =  λm2

Hence, wavelength will be halved.