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Structure of Atom · Question 55

AIPMT 2002MCQEasy📈 Topic in 38% papers+4−1
1

In hydrogen atom, energy of first excited state is -3.4 eV. Then find out K.E. of same orbit of hydrogen atom

  1. +3.4 eV
  2. +6.8 eV
  3. -13.6 eV
  4. +13.6 eV

Answer

(1) +3.4 eV

Solution

In a Bohr orbit, K.E. = 12 Ze24 0 r and P.E. = - Ze24 0 r

⇒ P.E. = -2 K.E.

⇒ Total energy E = K.E. + P.E. = -K.E.

∴ K.E. = -E = -(-3.4 eV) = +3.4 eV

Answer: (1)

AIPMT 2002

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2

23592U, nucleus absorbs a neutron and disintegrates into 13954Xe, 9438Sr and x so, what will be product x?

  1. 3-neutrons
  2. 2-neutrons
  3. -particle
  4. -particle
AIPMT 2002
3

The following quantum numbers are possible for how many orbitals : n = 3, l = 2, m = +2 ?

  1. 1
  2. 2
  3. 3
  4. 4
AIPMT 2001
4

If a abX species emits first a positron, then two and two particles and in the last, one is also emitted and gets converted to cdY species. So correct relation is

  1. c = a - 12, d = b - 5
  2. c = a - 5, d = b - 1
  3. c = a - 6, d = b - 0
  4. c = a - 4, d = b - 2
AIPMT 2001
5

For given energy, E = 3.03 10-19 Joules corresponding wavelength is (h = 6.626 10-34 J sec, c = 3 108 m/sec)

  1. 65.6 nm
  2. 6.56 nm
  3. 3.4 nm
  4. 656 nm
AIPMT 2000
6

Isoelectronic species are

  1. CO, CN-, NO+, C22-
  2. CO-, CN, NO, C2-
  3. CO+, CN+, NO-, C2
  4. CO, CN, NO, C2
AIPMT 2000

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