Atoms

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Question
CBSEENPH12039899

A certain mass of hydrogen is changed to helium the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u =931 MeV)

  • 2.67 MeV

  • 26.7 MeV

  • 6.675 MeV

  • 13.35 MeV

Solution

C.

6.675 MeV

Here comma space increment straight m equals 0.02866 space straight U
therefore comma
Energy space liberated
equals space fraction numerator 0.02866 space straight x space 931 over denominator 4 end fraction space equals space fraction numerator 26.7 over denominator 4 end fraction space MeV space equals space 6.675 space MeV

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Question
CBSEENPH12039826

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in the uniform magnetic field. What should be the energy of an alpha particle to describe a circle of the same radius in the same field.

  • 2 MeV

  • 1 MeV

  • 0.5 MeV

  • 4 MeV

Solution

B.

1 MeV

For space proton
straight r space equals space fraction numerator square root of 2 straight m left parenthesis KE right parenthesis end root over denominator qB end fraction
So
straight q space proportional to space square root of straight m left parenthesis KE right parenthesis end root

Hence space fraction numerator straight e over denominator 2 straight e end fraction space equals space square root of fraction numerator left parenthesis straight m subscript straight p right parenthesis left parenthesis 1 space Mev right parenthesis over denominator left parenthesis 4 straight m subscript straight p right parenthesis left parenthesis KE right parenthesis end fraction end root
1 fourth space equals fraction numerator 1 space MeV over denominator 4 KE end fraction

KE space equals 1 space MeV

Question
CBSEENPH12039838

An electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ1:λ2 emitted in the two cases is

  • 7/5

  • 27/20

  • 27/5

  • 20/7

Solution

D.

20/7

Here, for wavelength λ1
n1 = 4 and n2 =3
and for λ2, n1 = 3 and n2 = 2
We space have space hc over straight lambda space equals space minus 13.6 open square brackets fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction close square brackets
so comma space for space straight lambda subscript 1
rightwards double arrow space hc over straight lambda subscript 1 space equals space minus 13.6 open square brackets fraction numerator 1 over denominator left parenthesis 4 right parenthesis squared end fraction minus fraction numerator 1 over denominator left parenthesis 3 right parenthesis squared end fraction close square brackets
hc over straight lambda subscript 1 space equals space minus 13.6 open square brackets 7 over 144 close square brackets space.... space left parenthesis straight i right parenthesis space
simiarly comma space for space straight lambda subscript 2
rightwards double arrow space hc over straight lambda subscript 1 space equals space minus 13.6 open square brackets fraction numerator 1 over denominator left parenthesis 3 right parenthesis squared end fraction minus fraction numerator 1 over denominator left parenthesis 2 right parenthesis squared end fraction close square brackets
hc over straight lambda subscript 1 space equals space minus 13.6 open square brackets 5 over 36 close square brackets... space left parenthesis ii right parenthesis
Hence comma space from space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
straight lambda subscript 1 over straight lambda subscript 2 space equals space 20 over 7

Question
CBSEENPH12039849

An electron of a stationary hydrogen atom passes from the fifth energy level to ground level. The velocity that the atom acquired as a result of photon emission will be

  • fraction numerator 24 hR over denominator 25 space straight m end fraction
  • fraction numerator 25 hR over denominator 24 space straight m end fraction
  • fraction numerator 25 space straight m over denominator 24 space hR end fraction
  • fraction numerator 24 space straight m over denominator 25 space straight h space straight R end fraction

Solution

A.

fraction numerator 24 hR over denominator 25 space straight m end fraction Here comma space straight E subscript 5 space minus straight E subscript 1 space equals space hc over straight lambda
and space
fraction numerator straight R space hc over denominator 25 end fraction space equals space straight R space hc space equals space hc over straight lambda
24 over 25 straight R space equals space 1 over straight lambda
space straight P space equals straight h over straight lambda
straight v equals straight h over mλ
space equals space 24 over 25 Rh over straight m

Question
CBSEENPH12039804

Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given K= 9 x 109 constant, Z=2 and h (Planck's constant = 6.6 x 10-34 Js-1

  • 2.92 x 106 m/s

  • 1.46 x 106 m/s

  • 0.73 x 106 m/s

  • 3.0 x 108 m/s

Solution

B.

1.46 x 106 m/s

Energy of electron in the 3rd orbit of He+ is
straight E subscript 3 space equals space minus 13.6 space straight x space straight Z squared over straight n squared eV space equals space minus 13.6 space straight x space 4 over 3 squared eV
space equals space minus 13.6 space straight x space 4 over 9 straight x space 16 space straight x space 10 to the power of negative 19 end exponent space straight J
From space Bohr apostrophe straight s space model comma
straight E subscript 3 space equals negative KE subscript 3 space equals space minus 1 half straight m subscript straight e straight V squared
rightwards double arrow space 1 half space straight x space 9.1 space straight x space 10 to the power of negative 31 end exponent space straight x space straight v squared
space equals space minus 13.6 space straight x space 4 over 9 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent
straight v squared space equals space fraction numerator 136 space straight x space 16 space straight x space 4 space straight x space 2 straight x space 10 to the power of negative 11 end exponent over denominator 9 space straight x space 91 end fraction
or space straight v equals space 1.46 space straight x space 10 to the power of 6 space straight m divided by straight s

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