The manifestation of band structure in solids is due to
Heisenberg’s uncertainty principle
Pauli’s exclusion principle
Bohr’s correspondence principle
Boltzmann’s law
Which of the following transitions in a hydrogen atom emits of the highest frequency?
n = 1 to n = 2
n = 2 to n = 6
n = 2 to n = 1
n = 6 to n = 2
When electron jumps from n = 4 to n = 2 orbit, we get
Second line of Lyman series
Second line of Balmer series
Second line of Paschen series
An absorption line of Balmer series
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to wavelength of emitted radiation corresponds to the transition between
n = 3 to n = 2 states
n = 3 to n = 1 states
n = 2 to n = 1 states
n = 4 to n = 3 states
The spectrum obtained from a sodium vapour lamp is an example of
Band spectrum
Continuous spectrum
Emission spectrum
Absorption spectrum
An α-particle of energy 5 MeV is scattered through 180o by a fixed Uranium nucleus. The distance of the closest approach is of the order of
1 Å
10-10 cm
10-12 cm
10-15 cm
In terms of Bohr radius a0, the radius of the second Bohr orbit of hydrogen atom is given by
4 a0
8 a0
√2 a0
2 a0
If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
102 eV
Zero
3.4 eV
6.8 eV
Hydrogen atoms are excited from ground state to the principal quantum number 4. Then the number of spectral lines observed will be
3
6
5
2
When a hydrogen atom is raised from the ground state to an excited state
Potential energy decreases and kinetic energy increases
Potential energy increases and kinetic energy decreases
Both kinetic energy and potential energy decrease
