NEET Physics - New - Atoms and Nuclei

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Question - 1

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:

  • A 1
  • B 4
  • C 0.5
  • D 2

Question - 2

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength \(\lambda \). When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be :

  • A (16/25)\(\lambda \)
  • B (9/16)\(\lambda \)
  • C (20/7)\(\lambda \)
  • D (20/13)\(\lambda \)

Question - 3

Consider 3rd orbit of He+ (Helium) using nonrelativistic approach, the speed of electron in this orbit will be :

  • A 0.73 x 106 m/s
  • B 3.0 x 108 m/s
  • C 2.92 x 106 m/s
  • D 1.46 x 106 m/s

Question - 4

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:

  • A 5/27
  • B 3/23
  • C 7/29
  • D 9/3

Question - 5

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

  • A 1 MeV
  • B 0.5 MeV
  • C 4 MeV
  • D 2 MeV

Question - 6

Radioactive material 'N has decay constant '8\(\lambda \)' and material 'B' has decay constant '\(\lambda \)'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'N will be 1/e?

  • A 1/7\(\lambda \)
  • B 1/8\(\lambda \)
  • C 1/9\(\lambda \)
  • D 1/\(\lambda \)

Question - 7

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for hydrogen like ion. The atomic number Z of hydrogen like ion is :

  • A 4
  • B 1
  • C 2
  • D 3

Question - 8

The energy of a hydrogen atom in the ground state is - 13.6 eV. The energy of He+ ion in the first excited state will be :

  • A -13.6 eV
  • B - 27.2 eV
  • C - 54.4 eV
  • D - 6.8 eV

Question - 9

An alpha nucleus of energy 1/2mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to :

  • A 1/Ze
  • B \({ \nu }^{ 2 }\)
  • C 1/m
  • D 1/\({ \nu }^{ 2 }\)

Question - 10

A source S1 is producing, 1015 photons per second of wavelength 5000 \(\mathring { A } \). Another source 52 is producing 1.02 x 1015 photons per second of wavelength 5100\(\mathring { A } \). Then, (power of S2)/(power of S1) is equal to :

  • A 1.00
  • B 1.02
  • C 1.04
  • D 0.98
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