What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m =-1.
10
6
4
2
An atomic nucleus having low n/p ratio tries to find stability by
the emission of an α-particle
the emission of a positron
capturing an orbital electron (K-electron capture)
emission of β -particle
(32Ge76, 34Se76) and (14Si30, 16S32) are examples of
isotopes and isobars
isobars and isotones
isotones and isotopes
isobars and isotopes
98Cf246 was formed along with a neutron when an unknown radioactive substance was bomabrded with 6C12. The unknown substance is
91Pa234
90Th234
92U235
92U238
Based on equation certain conclusions are written. Which of them is not correct?
The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
Larger the value of n, the larger is the orbit radius
Equation can be used to calculate the change in energy when the electron changes orbit.
For n = 1 the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
The energy of an electron in first Bohr orbit of H-atom is-13.6 eV. The possible energy value of electron in the excited state of Li2+ is:
- 122.4 eV
30.6 eV
- 30.6 eV
13.6 eV
The representation of the ground state electronic configuration of He by box-diagram as
is wrong because it violates:
Heisenberg's uncertainty principle
Bohr's quantization theory of angular momenta
Pauli exclusion principle
Hund's rule
The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state)
Li2+
He+
H
H+
A radioactive atom emits two particles and one particle successively. The number of neutrons in the nucleus of the product will be:
X - 4 - Y
X - Y - 5
X - Y - 3
X - Y - 6
B.
X - Y - 5
An -emission reduces the atomic mass by 4 and atomic number by 2, whereas a -emission increases the atomic number by 1 with no change in atomic mass.
Number of neutrons = mass number - atomic number
= X - 8 - Y + 3
= X - Y - 5
1 mole of photon, each of frequency 2500 s-1, would have approximately a total energy of
10 erg
1 J
1 eV
1 MeV